Here is a more careful (EDIT: even more careful!) argument that gives an affirmative answer to the weaker version of the question (as stated in the edit to my previous post, I doubt that the stronger version is true).

The argument uses the following lemma, which ought to be known. If someone has a reference, please leave a comment.

Lemma:
Let $a_1,\dots, a_n$ be real numbers with each $a_i\geq 1$, and let $X_1,\dots,X_n$ be independent random variables, each uniform on $\pm 1$. Let $I$ be an interval of length $2r$. Then $$Pr(a_1X_1+\cdots+a_nX_n\in I) \leq \frac{1+r}{\sqrt{\pi n/2}}.$$

Proof:
Let $f(X)$ denote $a_1X_1+\cdots+a_nX_n$.
In the Boolean lattice of all assignments of $\pm 1$ to the variables $X_1,\dots,X_n$, consider a random walk starting from the point where all $X_i$'s are $-1$, and moving in $n$ steps to the point where they are all $+1$, in each step choosing uniformly and independently of the history a variable which is $-1$ and changing it to $+1$.

What is the expectation of the number $N(I)$ of steps of this walk at which $f(X) \in I$? On one hand, $N(I)\leq 1+r$, since $f(X)$ increases by at least 2 in each step.

On the other hand, the probability that the walk passes through any given point in the Boolean lattice is at least $2^{-n}\sqrt{\pi n/2}$ (this probability is minimized at the middle level(s) of the lattice, and the claim follows by well-known estimates of the central binomial coefficient). Therefore
$$EN(I) \geq \frac{\#\{X:f(X)\in I\}}{2^n}\cdot \sqrt{\pi n/2} = Pr(f(X)\in I) \cdot \sqrt{\pi n/2}.$$

It follows that $$Pr(f(X)\in I) \leq \frac{1+r}{\sqrt{\pi n/2}}. \qquad \square$$

As was explained in the earlier post, we can randomly choose $n$ pairs of opposite points $\{z_i, -z_i\}$, then find $z$ with $\left|P(z)P(-z)\right|=1$ given only this information, and finally fix the $z_i$'s by $n$ independent coin flips.

In order to apply the lemma, we want to have, before the coin flipping, $n/2$ pairs $z_i, -z_i$ making an angle of say at most $60$ degrees with $z, -z$, so that each of the $n/2$ corresponding coin flips determine the sign of a term of at least $\log 3$ in $\log\left|P(z)\right| - \log\left|P(-z)\right|$. Actually, after choosing the $n$ pairs $z_i, -z_i$, this a. a. s. holds for every $z$. The idea is to divide the circle into, say, 100 equally large sectors. With high probability, every pair of opposite sectors will contain at least $n/51$ pairs (as opposed to the expected number, $n/50$).

We now condition on the outcomes of the coin flips for the smaller terms (pairs $z_i, -z_i$ more or less orthogonal to $z$). The lemma above tells us that for any interval $I$ of length $4\alpha\sqrt{n}$, the probability that $\log\left|P(z)\right| - \log\left|P(-z)\right| \in I$ is at most $4\alpha/\sqrt{\pi} + O(1/\sqrt{n})$.

In particular, with probability at least $1-O(\alpha)$, the absolute value of $\log\left|P(z)\right| - \log\left|P(-z)\right|$ is at least $2\alpha\sqrt{n}$, and since $\log\left|P(z)\right|=-\log\left|P(-z)\right|$, $\max\left(\log\left|P(z)\right|, \log\left|P(-z)\right|\right)\geq \alpha\sqrt{n}$ as required.