Ah, I think I can stymy $M$ with 2 nonconstant advisors. Namely, let $A_1\left(n\right)=\frac{1}{2}-\frac{1}{n+3}$ and $A_2\left(n\right)=\frac{1}{2}+\frac{1}{n+3}$. We (setting up an adversarial $E$) precommit to setting $E\left(n\right)=0$ if $p\left(n\right)\ge A_2\left(n\right)$ and $E\left(n\right)=1$ if $p\left(n\right)\le A_1\left(n\right)$; now we can assume that $M$ always chooses $p\left(n\right)\in \left[A_1\right(n),A_2(n\left)\right]$, since this is better for $M$.

Now define $b_i^{\prime }\left(j\right)=\left|A_i\right(j)+E(j)-1|-\left|p\right(j)+E(j)-1|$ and $b_i\left(n\right)={\sum }_{j<n}{b}_i^{\prime }\left(j\right)$. Note that if we also define ${\mathrm{bad}}_i\left(n\right)={\sum }_{j<n}(\log \left|A_i\right(j)+E(j)-1|-\log \left|p\right(j)+E(j)-1|)$ then ${\sum }_{j<n}\left|2b_i\right(j)-{\mathrm{bad}}_i(j\left)\right|\le {\sum }_{j<n}\left(2A_1\right(j)-1-\log (2A_1\left(j\right)\left)\right))={\sum }_{j<n}O\left({(\frac{1}{2}-A_1(j\left)\right)}^2\right)$ is bounded; therefore if we can force $b_1\left(n\right)\to \infty$ or $b_2\left(n\right)\to \infty$ then we win.

Let's reparametrize by writing $\delta \left(n\right)=A_2\left(n\right)-A_1\left(n\right)=\frac{2}{n+3}$ and $q\left(n\right)=\frac{p\left(n\right)-A_1\left(n\right)}{\delta \left(n\right)}$, so that $b_i^{\prime }\left(j\right)=\delta \left(j\right)(|i-2+E(j\left)\right|-\left|q\right(j)-1+E(j\left)\right|)$.

Now, similarly to how $M$ worked for constant advisors, let's look at the problem in rounds: let $s_0=0$, and $s_n=\lfloor \exp (s_{n-1}-1)\rfloor +1$ for $n>0$. When determining $E\left(s_{n-1}\right),\dots ,E(s_n-1)$, we can look at $p\left(s_{n-1}\right),\dots ,p(s_n-1)$. Let $t_n=\lfloor b_2\left(s_n\right)-\frac{1}{n}\rfloor$. Let's set $E\left(s_{n-1}\right),\dots ,E(s_n-1)$ to 1 if ${\sum }_{j=s_{n-1}}^{s_n-1}\delta \left(j\right)(1-q(j\left)\right)\ge 1$; otherwise we'll do something more complicated, but maintain the constraint that $b_2\left(s_n\right)\ge b_2\left(s_{n-1}\right)-\frac{1}{n(n-1)}\ge t_{n-1}+\frac{1}{n}$: this guarantees that $t_n$ is nondecreasing and that ${liminf}_{j\to \infty }{b}_2\left(j\right)\ge {lim}_{n\to \infty }{t}_n$.

If $t_n\to \infty$ then $b_2\left(n\right)\to \infty$ and we win. Otherwise, let $t={lim}_{n\to \infty }{t}_n$, and consider $n$ such that $t_{n-1}=t$.

We have ${\sum }_{j=s_{n-1}}^{s_n-1}\delta \left(j\right)(1-q(j\left)\right)<1$. Let $J\subseteq \{s_{n-1},\dots ,s_n-1\}$ be a set of indices with $q\left(j\right)\ge q\left(j^{\prime }\right)$ for all $j\in J,j^{\prime }\notin J$, that is maximal under the constraint that ${\sum }_{j\in J}\delta \left(j\right)(1-q(j\left)\right)\le \frac{1}{n(n-1)}$; thus we will still have ${\sum }_{j\in J}\delta \left(j\right)(1-q(j\left)\right)\ge \frac{1}{n(n-1)}-\delta \left(s_{n-1}\right)$. We shall set $E\left(j\right)=0$ for all $j\in J$.

By the definition of $J$: $$\begin{array}{cc}\sum _{j\in J}{b}_1^{\prime }\left(j\right)& =\sum _{j\in J}\delta \left(j\right)q\left(j\right)\\ & \ge \sum _{j\in J}\delta \left(j\right)(1-q(j\left)\right)\frac{{\sum }_{j=s_{n-1}}^{s_n-1}\delta \left(j\right)q\left(j\right)}{{\sum }_{j=s_{n-1}}^{s_n-1}\delta \left(j\right)(1-q(j\left)\right)}\\ & \ge (\frac{1}{n(n-1)}-\delta (s_{n-1}\left)\right)\frac{{\sum }_{j=s_{n-1}}^{s_n-1}\delta \left(j\right)-1}{1}\\ & \ge (\frac{1}{n(n-1)}-\delta (s_{n-1}\left)\right)(2\log \left(\frac{s_n+3}{s_{n-1}+3}\right)-1)\\ & \ge 2\text{if}n\gg 0\end{array}$$

For $j^{\prime }\notin J$, we'll proceed iteratively, greedily minimizing $\left|{\sum }_{j^{\prime }=s_{n-1}}^j{1}_{j^{\prime }\notin J}\left(b_1^{\prime }\right(j^{\prime }),b_2^{\prime }(j^{\prime }\left)\right)\right|$. Then: $$\begin{array}{cc}\underset{s_{n-1}\le j<s_n}{min}\sum _{j^{\prime }=s_{n-1}}^j{1}_{j^{\prime }\notin J}{b}_1^{\prime }\left(j^{\prime }\right)& \ge -\sqrt{\sum _{j=s_{n-1}}^{s_n-1}\delta (j{)}^2}\\ & =-2\sqrt{\sum _{j=s_{n-1}+3}^{s_n-2}\frac{1}{j^2}}\\ & \ge -2\sqrt{\sum _{j=s_{n-1}+3}^{s_n-2}(\frac{1}{j-1}-\frac{1}{j})}\\ & \ge -\frac{2}{\sqrt{s_{n-1}+2}}\\ & \ge -1\text{if}n\gg 0\end{array}$$

Keeping this constraint, we can flip (or not flip) all the $E\left(j^{\prime }\right)$s for $j^{\prime }\notin J$ so that ${\sum }_{j^{\prime }=s_{n-1}}^{s_n-1}{1}_{j^{\prime }\notin J}{b}_2^{\prime }\left(j^{\prime }\right)>0$. Then, we have $b_2\left(s_n\right)\ge b_2\left(s_{n-1}\right)-\frac{1}{n(n-1)}$, $b_1\left(s_n\right)-b_1\left(s_{n-1}\right)={\sum }_{j=s_{n-1}}^{s_n}(1_{j\in J}+1_{j\notin J})b_1^{\prime }\left(j\right)\ge 2-1=1$ if $n\gg 0$, and for $s_{n-1}\le j\le s_n$, $b_1\left(j\right)\ge b_1\left(s_{n-1}\right)+{\sum }_{j^{\prime }=s_{n-1}}^{j-1}{1}_{j^{\prime }\notin J}{b}_1^{\prime }\left(j^{\prime }\right)\ge b_1\left(s_{n-1}\right)-1$ if $n\gg 0$.

Therefore, $b_1\left(j\right)\to \infty$, so we win.

I don't yet know whether I can extend it to two nonconstant advisors, but I do know I can extend it to a countably infinite number of constant-prediction advisors. Let $(P_i{)}_{i=0,\dots }$ be an enumeration of their predictions that contains each one an infinite number of times. Then:

def M(p, E, P):
    prev, this, next = 0, 0, 1
    def bad(i):
        return sum(log(abs((E[k] + P[i] - 1) /
                           (E[k] + p[k] - 1)))
                   for k in xrange(prev))
    for k in xrange(this, next): p[k] = 0.5
    prev, this, next = this, next, floor(exp(next - 1)) + 1

    for i in xrange(0, Inf):
        for k in xrange(this, next): p[k] = P[i]
        prev, this, next = this, next, floor(exp(next - 1)) + 1

bad(i) is now up to date through E[:this], not just E[:prev]

        bound = 2 * bad(i)
        for j in xrange(0, Inf):
            if P[j] == P[i]: continue
            flip = P[j] < P[i]
            p1, p2 = abs(P[i] - flip), abs(P[j] - flip)
            for k in xrange(this, next): p[k] = abs(p1 - flip)
            prev, this, next = this, next, floor(exp(next - 1)) + 1
            
            if bad(i) <= 0: break
            while bad(i) > 0 and bad(j) > 0:
                # won't let bad(i) surpass bound
                eps = (bound - bad(i)) / 2 / abs(1 - p1 - flip) / (next - this)

This is just for early iterations of the inner loop; in the limit, eps should be just enough for bad(i) to go halfway to bound if we let p = abs(p1 + eps - flip):

                while eps >= 1 - p1 or
                      bound <= bad(i) + (next - this) *
                        log((1 - p1) / (1 - p1 - eps)):
                    eps /= 2
                for k in xrange(this, next): p[k] = abs(p1 + eps - flip)
                prev, this, next = this, next, floor(exp(next - 1)) + 1

                for k in xrange(this, next): p[k] = abs(p1 - flip)
                # this is where the P[i] + d * eps affects bad(i)

Consider $q=\frac{\log (1-p1)-\log (1-p2)}{\log (1-p1)-\log (1-p2)+\log \left(p2\right)-\log \left(p1\right)}$. This $q$ is the probability between p1 and p2 such that if E[k] is chosen with probability $|q-flip|$ then that will have an equal impact on bad(i) and bad(j). Now consider some $q^{\prime }$ between p1 and $q$. Every iteration where $\mathrm{mean}\left(\left|E[prev:this]-flip\right|\right)\le q^{\prime }$ will decrease bad(j) by a positive quantity that's at least linear in this-prev, so (at least after the first few such iterations) this will exceed $prev\cdot -\log ({max}_{k:P_k\text{has been reached}}max(P_k,1-P_k\left)\right)>bad\left(j\right)$, so it will turn bad(j) negative. If this happens for all j then M cannot be bad for E. If it doesn't, then let's look at the first j where it doesn't. After a finite number of iterations, every iteration must have $\mathrm{mean}\left(\left|E[prev:this]-flip\right|\right)>q^{\prime }$. However, this will cause bad(i) to decrease by a positive quantity that's at least proportional to bound - bad(i); therefore, after a finite number of such iterations, we must reach $bad\left(i\right)<0$. So if M is bad for E then for each value of i we will eventually make $bad\left(i\right)<0$ and then move on to the next value of i. This implies M is not bad for E.

Emboldened by this, we can also consider the problem of building an $M$ that isn't outperformed by any constant advisor. However, this cannot be done, according to the following handwavy argument:

Let $q$ be some incompressible number, and let $E\left(i\right)\stackrel{iid}{\sim }\mathrm{Bern}\left(q\right)$. When computing $p\left(n\right)$, $M$ can't do appreciably better than Laplace's law of succession, which will give it standard error $\sqrt{\frac{q(1-q)}{\log \left(n\right)}}$, and relative badness $\sim \frac{q(1-q)}{\log \left(n\right)}(\frac{1}{q}+\frac{1}{1-q})=\frac{1}{\log \left(n\right)}$ (relative to the $q$-advisor) on average. For $i\le n$, and $n\gg 0$, the greatest deviation of the badness from the ${\sum }_{j=2}^i\frac{1}{\log \left(j\right)}\ge \frac{i-1}{\log \left(i\right)}$ trend is $\approx \sqrt{2n\log \log \left(n\right)q(1-q)}$ (according to the law of the iterated logarithm), which isn't enough to counteract the expected badness; therefore the badness will converge to infinity.