Chebyshev polynomials suggest the answer should be $\frac{1}{2^{n-1}}$. I believe interpolation should be sufficient to prove this. Also I think this is in PFTB. EDIT: I have found the problem, it is due to Marius Cavachi and was on a Romanian Olympiad. The answer is actually unity and not $\frac{1}{2^{n-1}}$.

Indeed, the answer is $1$. Obviously, if $p(z) = z^n$ then $\max_{|z| = 1} |p(z)| = 1$. Thus it suffices to show that $\max_{|z| = 1} |p(z)| \ge 1$ for every $p \in P$. Suppose fsoc that there exists $p \in P$ such that for all $z$ with $|z|=1$ we have $$|p(z)| < 1 = |z^n |.$$By Rouche's theorem, $z^n - p(z)$ then has $n$ roots in the unit disc, counted with multiplicity. However, this is impossible since $z^n - p(z)$ is a polynomial of degree $\le n-1$. I don't think this problem is suitable for high school olympiads.

MathisWow wrote: Let $n$ be a positive integer and let $P$ be the set of monic polynomials of degree $n$ with complex coefficients. Find the value of \[ \min_{p \in P} \left \{ \max_{|z| = 1} |p(z)| \right \} \] This is just a very special case of Problem 7 from IMC 2021. See here.

Gryphos wrote: Indeed, the answer is $1$. Obviously, if $p(z) = z^n$ then $\max_{|z| = 1} |p(z)| = 1$. Thus it suffices to show that $\max_{|z| = 1} |p(z)| \ge 1$ for every $p \in P$. Suppose fsoc that there exists $p \in P$ such that for all $z$ with $|z|=1$ we have $$|p(z)| < 1 = |z^n |.$$By Rouche's theorem, $z^n - p(z)$ then has $n$ roots in the unit disc, counted with multiplicity. However, this is impossible since $z^n - p(z)$ is a polynomial of degree $\le n-1$. I don't think this problem is suitable for high school olympiads. one nice solution!

One could also mimic the proof for the $[-1,1]$ interval case: if $\omega_1, \omega_2, \cdots \omega_n$ are the $n$-th roots of unity, then by Lagrange interpolation the leading coefficient of $p(z)$ is equal to $\sum_{j=1}^n \frac{p(\omega_j)}{\prod_{k\neq j} (\omega_j - \omega_k)}$. Since the absolute value of the denominators is $n$, triangle inequality gives $1=\bigg| \sum_{j=1}^n \frac{p(\omega_j)}{\prod_{k\neq j} (\omega_j - \omega_k)} \bigg| \leq \sum_{j=1}^n \frac{|p(\omega_j)|}{n}$. Thus at least one of $|p(\omega_1)|, |p(\omega_2)|, \cdots |p(\omega_n)|$ is $\geq 1$.