If we consider $\pmod{16}$, we have $$x_1^4+x_2^4+\cdots+x_{4}^{14} \equiv 0, 1, \ldots, 14 \pmod{16}$$ since $x_i^4 \equiv 0, 1 \pmod{16}$. Now if we try to do the same thing to RHS we get $$2016^3 \equiv 0 \pmod{16} \implies 2016^3 - 1 \equiv -1 \equiv 15 \pmod{16}$$ but the RHS can only get atmost $14$, so we get no solution.