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.
Here is a detailed and self-contained proof for general $n$, which also covers the "equality" case. It is based on Fedja's post, but it only uses (a variant of) the Gauss-Lucas theorem once.
Let $a_0,\dotsc,a_{n-1}\in\mathbb{C}$ be coefficients such that
$$p(z):=(z-z_1)\dotsb(z-z_n)=(z-1)^n+\sum_{k=0}^{n-1}a_k z^k,$$
and assume that $(z-1)^{n-1}$ does not divide the left-hand side. Then, the $k$-sum on the right-hand side is a polynomial of degree $n-1$, because
$$\Re(a_{n-1}) = \Re\left(n-\sum_{j=1}^n z_j\right)=n-\sum_{j=1}^n\Re(z_j)>0,$$
and we need to prove that $|a_{n-1}|>|a_0|$.
Introducing the differential operator
$$Df(z):=(1-z)f'(z)+nf(z),$$
we claim that every root of the polynomial $Dp(z)$ lies in the closed unit disk. Indeed, assume for a contradiction that $Dp(z)=0$ and $|z|>1$.
Then $p'(z)/p(z)=n/(z-1)$, that is,
$$\frac{1}{n}\sum_{j=1}^n \frac{1}{z-z_j}=\frac{1}{z-1}.$$
Let $K$ be the image of the closed unit disk under the Möbius transformation $s\mapsto 1/(z-s)$. Using $|z|>1$, we see that $K$ is a closed disk containing $1/(z-1)$ on its boundary. By the previous display, this boundary point is a convex linear combination of the points $1/(z-z_j)$, which also lie in $K$. This forces that all the $z_j$'s are equal to $1$, contrary to our assumption. So the claim is proved.
Now observe that the polynomial $Dp(z)$ is the same as
$$D\left(\sum_{k=0}^{n-1}a_k z^k\right)=a_{n-1}z^{n-1}+\sum_{k=0}^{n-2}((n-k)a_k+(k+1)a_{k+1})z^k.$$
We proved that this polynomial has all its roots in the closed unit disk, therefore
$$|(n-k)a_k+(k+1)a_{k+1}|\leq\binom{n-1}{k}|a_{n-1}|,\qquad k\in\{0,\dotsc,n-2\}.$$
Applying the triangle inequality and re-arranging, we get that
$$\frac{n-k}{k+1}|a_k|\leq|a_{k+1}|+\frac{1}{k+1}\binom{n-1}{k}|a_{n-1}|.\tag{$\ast$}$$
From here we derive by induction that
$$\binom{n}{k}|a_0|\leq |a_k|+\binom{n-1}{k-1}|a_{n-1}|,\qquad k\in\{1,\dotsc,n-1\}.$$
In particular, the special case $k=n-1$ tells us that $n|a_0|\leq n|a_{n-1}|$.
To finish the proof, we only need to show that some of our inequalities are strict, so that the final inequality is strict as well. Let us assume that equality holds in each of our inequalities. Then the roots of $Dp(z)$ are equal to a single $w$ on the unit circle, so that
$$(n-k)a_k+(k+1)a_{k+1}=\binom{n-1}{k}a_{n-1}(-w)^{n-1-k},\qquad k\in\{0,\dotsc,n-2\}.$$
In particular, the case $k=n-2$ yields
$$\frac{2}{1-n}a_{n-2}=(1+w)a_{n-1}.$$
However, we have equality in $(\ast)$ for $k=n-2$, whence $|1+w|=2$, which forces $w=1$. But then the recursion yields that
$$a_k=\binom{n-1}{k}(-1)^{n-1-k}a_{n-1},$$
i.e.,
$$p(z)=(z-1)^n+a_{n-1}(z-1)^{n-1}.$$
This contradicts our initial assumption, and we are done.
Added. As Malkoun pointed out in the comments, $(\ast)$ also implies the inequalities
$$|a_k|\leq\binom{n-1}{k}|a_{n-1}|,\qquad k\in\{0,\dotsc,n-1\}.$$
Hence, renaming $k$ to $n-k$, we get the following generalization of the original inequality:
$$\left|\sum_{1\leq j_1<\dotsb< j_k\leq n}z_{j_1}\dotsb z_{j_k}-\binom{n}{k}\right|\leq\binom{n-1}{k-1}|z_1+\dotsb+z_n-n|,\qquad k\in\{1,\dotsc,n\}.$$
Best Answer
The answer is no. E.g., let $n=3$ and $z_j=e^{i(j-1)2\pi/3}$ for $j=1,2,3$. Then $f(z)>|z|+15/100>|z|$ if $|z|\le1$.
This counterexample generalizes to any $n\ge3$. Indeed, take any $n\ge3$ and let $z_j=e^{i(j-1)2\pi/n}$ for $j=1,\dots,n$. Then $$f(z)=\frac1n\,\sum_{j=1}^n f_j(z),$$ where $f_j(z):=|z-z_j|$. Note that for each $j$ the function $f_j$ is convex and, moreover, $f_j$ is strictly convex on any straight line not through the point $z_j$. Since $n\ge3$, there is no straight line passing through all the points $z_1,\dots,z_n$. So, the function $f=\frac1n\,\sum_{j=1}^n f_j$ is strictly convex.
Also, for any real $a$ and $b$, the derivative of $f$ at $0$ along the vector $(a,b)=a+ib$ is $$\begin{aligned}&\frac d{ds}\,f(s(a+ib))\Big|_{s=0} \\ &=-\frac1n\,(a,b)\cdot\sum_{j=1}^n \Big(\cos\frac{(j-1)2\pi}n,\sin\frac{(j-1)2\pi}n\Big) \\ &=-\frac1n\,(a,b)\cdot(0,0)=0, \end{aligned}$$ where $\cdot$ denotes the dot product.
Thus, the strictly convex function $f$ has a unique minimum at $0$, and the minimum value of $f$ is $1$. That is, $f(0)=1<f(z)$ for all $z$ such that $0<|z|\le1$.
So, for all $z$ such that $0<|z|\le1$, we have $f(z)>1\ge|z|$ and hence $f(z)\ne|z|$. Also, $f(0)=1\ne0$. We conclude that there is no $z$ such that $|z|\le1$ and $f(z)=|z|$, which proves the claim.
On the other hand, the answer is yes for $n=1$ and $n=2$.