Solution:
Note that $\frac{1}{16} f(2x) = x^{4} + 7x^{3} + 13x^{2} + 7x + 1$. Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1) = -1$ and $f(-2) = 16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$:
$$\left|z_{\sigma(1)} z_{\sigma(2)} + z_{\sigma(3)} z_{\sigma(4)}\right| \leq \frac{1}{2}\left(z_{\sigma(1)} z_{\sigma(2)} + z_{\sigma(3)} z_{\sigma(4)} + z_{\sigma(4)} z_{\sigma(3)} + z_{\sigma(2)} z_{\sigma(1)}\right)$$
Let the roots be ordered $z_{1} \leq z_{2} \leq z_{3} \leq z_{4}$, then by rearrangement the last expression is at least:
$$\frac{1}{2}\left(z_{1} z_{4} + z_{2} z_{3} + z_{3} z_{2} + z_{4} z_{1}\right)$$
Since the roots come in pairs $z_{1} z_{4} = z_{2} z_{3} = 4$, our expression is minimized when $\sigma(1) = 1, \sigma(2) = 4, \sigma(3) = 3, \sigma(4) = 2$ and its minimum value is $8$.