Lagrangian interpolation at Chebyshev points - estimate on coefficients in monomic basis

First, let us fix some Notation:

Let $n\in\mathbb{N}$ and $x_i=\cos(\tfrac{(i+1/2)\pi}{(n+1)})$, $i=0,\dots,n$, be the Chebyshev points. Let \begin{align}L_i(x)={\displaystyle\prod_{\substack{0\leq j\leq n\\i\neq j}}}\frac{x-x_j}{x_i-x_j}, \end{align} $i=0,\dots,n$, be the Lagrange polynomials. Now let $f\in C^{\infty}([-1,1],\mathbb{R})$ with $\sup_{x\in[-1,1]}|f(x)|\leq 1$. We can now consider the interpolating polynomial given by \begin{align} P_n(x)=\sum_{i=0}^n f(x_i)L_i(x). \end{align} Now to the question:

By definition of the $L_i$ we know that $P_n$ is a polynomial of degree $n$, i.e. there exists coefficients $c_{n,j}$, $j=0,\dots,n$, such that \begin{align} P_n(x)=\sum_{j=0}^n c_{n,j}x^j. \end{align}
We are looking for an estimate on the size of the coefficients $c_{n,j}$, specifically we would like something like \begin{align} \max_{j=0,\dots,n}|c_{n,j}|\leq \pi(n) \end{align} for some polynomial $\pi$. (Or alternatively an argument of why this is not possible).

As we are not all that familiar with this topic, we have been trying find such a result in the literature. While there is huge amount of work on Lagrange interpolation and Chebyshev polynomials, this kind of estimate does not seem to be of interest for the usual applications. I would, however, find it surprising if no one had ever considered this question. If someone could point us to an answer, we would be very grateful.

A bound of the kind you're looking for is not possible, and it doesn't even matter what interpolation points you use (much less which particular polynomial basis you choose to work with, since that has no effect on the $c_{n,j}$ anyway).
For a given value of $n$, let $f$ be (for instance) the Chebyshev polynomial of the first kind, $T_n$. Then the interpolation will be exact, so also $P_n = T_n$. But the largest coefficient of $T_n$ in the monomial basis is $2^n$, as can be easily proved using the standard recursive formula for these polynomials. Of course, $2^n$ grows faster than any polynomial of $n$.