Let's denote $\Pi_n$ the vector space of polynomial maps on I:=[-1,1] of degree less than or equal to n. The optimization problem you quoted is equivalent to:

**Problem(2):** Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $R^2$ with the max norm). The magic of polynomials is that the solution is always unique, for *any* continuous function f. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial q such that |f(x)-q(x)| attains the maximum value in at least n points, with alternate sign.
(Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$  is the unique monic polynomial with all real simple zeros and that reaches the values 1 and -1 alternately between consecutive zeros. With a bit more works one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).