Given $\vec{a} \in \mathbb{R}^n$ and $b \in \mathbb{R}$ consider the function $f(x) = Th[\vec{a}.\vec{x}+b]$ on $x \in \{-1,1\}^n$ such that the ``threshold function (Th)" gives $1$ when the argument is non-negative and $0$ when the argument is negative.

Given this and an $\epsilon >0$ how low degree a polynomial $p$ can one find such $p(x) = f(x)$ on at least $1-\epsilon$ fraction of the vertices of the hypercube?