Note that $x^{12}$ is a convex function of $x$ so the maximum value of the expression is obatined if all the numbers belong to $\{\frac{-1}{\sqrt{3}},\sqrt{3}\}$ except one(Note that all of the $x_i$'s being either one of the two values is not possible).Let $1996-k$ of the numbers be $\sqrt{3}$ and $k$ of the numbers be $\frac{-1}{\sqrt{3}}$.The remaining number then will be $-318\sqrt{3}+\frac{k}{\sqrt{3}}-(1996-k)\sqrt{3}$.This number must also lie between $\frac{-1}{\sqrt{3}}$ and $\sqrt{3}$ so we obtain the inequality $6941 \le 4n \le 6945$.Thus the only possibility is $k=1736$ so the required maximum value is $\boxed{1736*3^{-6}+260*3^6+(\frac{4}{3})^6}$.