Solution:
Answer: $-\frac{7}{16}$

For convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \Phi_{14}(x)$, which is suggestive. Indeed, consider $\omega$ a primitive $14$-th root of unity; since $\omega^{7}=-1$, we have $4 \omega^{10}=-4 \omega^{3}$, $-7 \omega^{9}=7 \omega^{2}$, and so on, so that
$$P(\omega)=12\left(\omega^{6}-\omega^{5}+\cdots+1\right)=12 \Phi_{14}(\omega)=0$$
Dividing, we find
$$P(x)=\Phi_{14}(x)\left(4 x^{4}-3 x^{3}-2 x^{2}-3 x+4\right)$$
This second polynomial is symmetric; since $0$ is clearly not a root, we have
$$4 x^{4}-3 x^{3}-2 x^{2}-3 x+4=0 \Longleftrightarrow 4\left(x+\frac{1}{x}\right)^{2}-3\left(x+\frac{1}{x}\right)-10=0$$
Setting $y=x+1/x$ and solving the quadratic gives $y=2$ and $y=-5/4$ as solutions; replacing $y$ with $x+1/x$ and solving the two resulting quadratics give the double root $x=1$ and the roots $(-5 \pm i \sqrt{39})/8$ respectively. Together with the primitive fourteenth roots of unity, these are all the roots of our polynomial.

Explicitly, the roots are
$$e^{\pi i / 7},\ e^{3 \pi i / 7},\ e^{5 \pi i / 7},\ e^{9 \pi i / 7},\ e^{11 \pi i / 7},\ e^{13 \pi i / 7},\ 1,\ (-5 \pm i \sqrt{39}) / 8$$
The sum of squares of the roots of unity (including $1$) is just $0$ by symmetry (or a number of other methods). The sum of the squares of the final conjugate pair is $\frac{2\left(5^{2}-39\right)}{8^{2}}=-\frac{14}{32}=-\frac{7}{16}$.