Solution:

The answer is $\mathbf{(E)}$. We have $x^{6}-16 x^{4}+16 x^{2}-1=\left(x^{2}-1\right)\left(x^{4}-15 x^{2}+1\right)$ (all the roots are real), so if $z$ is a root different from $\pm 1$, we have $z^{4}=15 z^{2}-1$, $z^{6}=15 z^{4}-z^{2}=224 z^{2}-15$. Denoting by $z_{1}, \ldots, z_{4}$ these latter roots, we have:
$$\sum_{j=1}^{4} z_{j}=0, \quad \sum_{1 \leq i<j \leq 4} z_{i} z_{j}=-15$$
$$\begin{gathered} \sum_{i=1}^{4} z_{i}^{2}=\left(\sum_{j=1}^{4} z_{j}\right)^{2}-2 \sum_{1 \leq i<j \leq 4} z_{i} z_{j}=30, \\ \sum_{i=1}^{4} z_{i}^{6}=224 \sum_{i=1}^{4} z_{i}^{2}-60=6660, \end{gathered}$$
now reintroducing into the count the roots $\pm 1$ we find that the correct answer is 6662. (D'Aurizio)