Let $T_{n}(x)=\sum_{\nu=0}^{n}t_{n,\nu}x^{\nu}$ denote the Chebyshev polynomial (of the first kind) of degree $n$ and let $x_{n,\nu}=\cos\nu\pi/n$ for $0\leq\nu\leq n$.

The answer follows from the following two results :

1) Let $f$ be a polynomial (possibly complex), of degree at most $n$, such that
$$|f(x_{n,\nu})|\leq1,\qquad 0\leq\nu\leq n.$$
Then
$$|a_{n-2\mu}|\leq|t_{n,n-2\mu}|,\qquad0\leq\mu\leq n/2,$$
which becomes an equality for $f=T_{n}$.

2) Let $f$ be a polynomial (possibly complex), of degree at most $n$, such that
$$|f(x_{n-1,\nu})|\leq1,\qquad 0\leq\nu\leq n-1.$$
Then
$$|a_{n-2\mu-1}|\leq|t_{n-1,n-2\mu-1}|,\qquad0\leq\mu\leq (n-1)/2,$$
which becomes an equality for $f=T_{n-1}$.

These are Theorems 16.3.1 and 16.3.2 in :

Rahman, Q. I.; Schmeisser, G. Analytic theory of polynomials. London
Mathematical Society Monographs. New Series, 26. The Clarendon Press,
Oxford University Press, Oxford, 2002.

The proof of the first assertion consists in considering a one-parameter family of polynomials constructed from $f$ and $T_{n}$ and depending on a parameter $\theta\in(-1,1)$, and then using Descartes' rule of signs. The second assertion is a simple consequence of the first one.