For a polynomial $f$, write$$\|f\|=\sup_{-1\leq x\leq 1}|f(x)|.$$It is a classical result of A. Markov that $\|f'\|\leq (\deg f)^2\|f\|$.
Question. Is there a simple proof of the weaker result that, for some $C>0$,$$\|f'\|\leq Cn^2\|f\|$$for any polynomial $f$ of degree $n$?
(In fact, I would be happy with any bound of the form $\|f'\|\lesssim n^{O(1)}\|f\|$.)
Background
The Markov brothers' inequality states that, for each $k\geq 1$ and each polynomial $f$ of degree $n$,$$\|f^{(k)}\|\leq\frac{n^2(n^2-1)\cdots(n^2-(k-1)^2)}{1\cdot 3\cdot 5\cdots (2k-1)}\|f\|\tag{$*$}.$$This constant is optimal for any choice of $n$ and $k$, and is achieved by the Chebyshev polynomial$f=T_n$.
It seems that most proofs of this precise inequality are rather painful; the original proof due to V. Markov in the $k>1$ case was about 110 pages. Many proofs are collected in Shadrin's wonderful article "Twelve proofs of the Markov inequality". However, if one only wants a bound of the rough shape $\|f^{(k)}\|\lesssim n^{2k}\|f\|$, it suffices to apply the $k=1$ result along with induction on $k$.
Shadrin mentions that the standard modern proof of the $k=1$ case goes via Bernstein's inequality for trigonometric polynomials, which is enough to prove that $|f'(x)|\leq n^2\|f\|$ for $x$ not too close to $\pm 1$, and to use some different trigonometric-polynomial machinery to treat the other values of $x$. While this is certainly nice, it isn't that simple, and it requires some nontrivial machinery (a bit of complex analysis, explicit arguments about Chebyshev polynomials). Markov's original argument (at least, how I understand it from Shadrin's paper) is of the same form, in the sense of first proving Bernstein's inequality and then splitting into cases. There are many other approaches discussed by Shadrin, but I was not able to understand any of them as especially simple in the $k=1$ case.
Most of the literature surrounding these inequalities seems to be focused on achieving the precise constant in $(*)$, and I was not able to find any work discussing simplified approaches for weaker statements (nor any substantially different approach for the $k=1$ case).
I am aware of a very cute argument showing that, for each degree $n$, there is some constant $C_n$ for which $\|f'\|\leq C_n\|f\|$ for all $f$ of degree at most $n$ (namely: $f\mapsto \|f\|$ and $f\mapsto \|f'\|$ are both norms on the space $\mathbb R[x]_{\deg\leq n}\cong\mathbb R^{n+1}$, and so they are equivalent). However, I do not know a way to make such a topological argument quantitative.
Why do I care? The Markov brothers' inequality is an essential intermediate step in an argument I need to present. The constant factors are not so important in this application (only the polynomial shape of the bound), but it would be nice to be able to explain succinctly why such an inequality ought to be true.