It seems that your question is an open problem, but there are partial answers.
Let's fix natural number $n$, and $1. By $T_{n,p}$ we denote polynomial of degree $n$ with leading coefficient equa to $1$ with minimal norm in $L_p([-1,1])$.
It is known that [1]
- $T_{n,1}$ is the Chebyshev polynomial of the second kind, so $ T_{n,1}(x)=U_n(x)=\frac{\sin((n+1)\arccos(x))}{\sin(\arccos(x))} $
- $T_{n,2}$ is the Legendre polynomial, so $ T_{n,2}(x)=L_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n $
- $T_{n,\infty}$ is the Chebyshev polynomial of the first kind, so $ T_{n,\infty}(x)=T_n(x)=\cos(n\arccos(x)) $
However we can force Chebychev polynomials to minimize weighted norms in $L_p([0,1])$ [2]. In fact
$T_n(x)$ have the smallest norm $ \Vert f \Vert=\left(\int\limits_{-1}^1(1-x^2)^{-1/2}|f(x)|^p dx\right)^{1/p} $ among polynomials of degree $n$ with leading coefficient $1$.
$U_n(x)$ have the smallest norm $ \Vert f \Vert=\left(\int\limits_{-1}^1(1-x^2)^{(p-1)/2}|f(x)|^p dx\right)^{1/p} $ among polynomials of degree $n$ with leading coefficient $1$.