Assume $f(x)\in\Bbb{R}[x]$ is a polynomial of degree $n$.
Question. If $\int_{-1}^1f^2(x)\,dx=1$, is it true that $$\vert f(x)\vert\leq \frac1{\sqrt2}(n+1), \qquad \text{for $\vert x\vert\leq1$}\,\,\,?$$
Assume $f(x)\in\Bbb{R}[x]$ is a polynomial of degree $n$.
Question. If $\int_{-1}^1f^2(x)\,dx=1$, is it true that $$\vert f(x)\vert\leq \frac1{\sqrt2}(n+1), \qquad \text{for $\vert x\vert\leq1$}\,\,\,?$$
This is problem VI.103 in volume 2 of Polya and Szego, where they also characterize the extremal polynomials.
Inequalities between different $L^{p}$ norms of the type $\|q_{n}\|_{p}\leq C_{n}\|q_{n}\|_{q}$, $0<q<p\leq\infty$, where $q_{n}$ is a polynomial of degree $n$ are sometimes called Nikolskii's inequalities, see p.102 of
R.A. DeVore; G.G. Lorentz, Constructive approximation, Springer-Verlag, Berlin, 1993.
For the norm $$\|f\|_{p}=\left(\int_{a}^{b}|f|^{p}dx\right)^{1/p},$$ on an interval $[a,b]$, one has $$\|q_{n}\|_{p}\leq(2(q+1)n^{2}/(b-a))^{1/q-1/p}\|q_{n}\|_{q}.$$ The proof essentially uses Markov inequality. For $[a,b]=[-1,1]$, $q=2$ and $p=\infty$, it would give $$\|q_{n}\|_{\infty}\leq\sqrt{3}n\|q_{n}\|_{2}.$$ About the Legendre polynomials discussed in the comments, it is known that the sum $\sum_{k=0}^{n}P_{k}^{2}(x)$, that is the reciprocal of the so-called Christoffel function, behaves, for $n$ large, like $n/(\pi\sqrt{1-x^{2}})$ for $x\in(-1,1)$, and behaves like $Cn^{2}$, $C$ some constant, at the endpoints $\pm1$.
This may not be of interest to you, but the weaker bound $|f(x)|\le (2+o(1))n$ has an easy proof, based on Bernstein's inequality $$ |f'(x)| \le \frac{n}{\sqrt{1-x^2}} \|f\|_{\infty} . \quad\quad\quad\quad (B) $$ This puts limits on how fast $f$ can decay near its maximum, and the worst case scenario occurs when the maximum is at $\pm 1$ (let's say at $x=1$). If we write $(1-x^2)^{-1/2}=(x+1)^{1/2}(1-x)^{1/2}$ and approximate the first factor by $\sqrt{2}$ near $x=1$, then integrating (B) gives that $|f(x)|\ge n\|f\|_{\infty}\sqrt{2(x-1+1/2n^2)}$ on $1-1/2n^2\le x\le 1$. This makes a contribution $\ge \|f\|_{\infty}^2/(4n^2)$ to the $L^2$ norm, so $\|f\|_{\infty}\le 2n$, as claimed. (Doing it properly produces the $o(1)$ term.)
The constant $2$ could be improved somewhat since (B) also holds with $n^2$ in place of $n/\sqrt{1-x^2}$, but I don't think I could get close to the conjectured optimal value $1/\sqrt{2}$ in this way.