Bounding a sum involving a $\Re((z\zeta)^N)$ term

The calculations were too awful and I haven't managed to find a simple form yet, but along with your first equality ($\sum \limits_{v=1}^{\frac{k-1}{2}} \sec^2 (\frac{2\pi v}{k} ) = \frac{1}{2} (k^2 - 1)$ ), you have also

$\sum \limits_{v=1}^{\frac{k-1}{2}} \frac{1}{\cos \big( \frac{2 \pi v}{k}\big)} = (-1)^\frac{k-1}{2} 2 \lfloor \frac{k-1}{4} \rfloor$.

And you can prove that $\sum \limits_{v=0}^{k-1} \cos ^j \big( \frac{2\pi v}{k} \big) = \frac{k}{2^j} \sum \limits_{\substack{0 \leqslant p \leqslant j\\ k \mid 2p - j}} \binom{j}{p}$.

Thus $$\sum \limits_{v=1}^{\frac{k-1}{2}} \cos ^j \big( \frac{2\pi v}{k} \big) = \frac{1}{2} \Big( \frac{k}{2^j} \sum \limits_{\substack{0 \leqslant p \leqslant j\\ k \mid 2p - j}} \binom{j}{p} - 1\Big)$$

And you can write your sum as a sum of sums of these types, which you can rewrite with the previous formula (but as I said, for now I haven't been able to do some real simplifications)