I edited the post and title.
How do we see that given $$ Z= \oint \frac{d \epsilon}{2\pi i} (z\bar z)^{-\epsilon} \frac{\pi^4 \sin 5\pi \epsilon}{\sin^5 \pi \epsilon} \left|\sum_{k=0}^\infty (-z)^k \frac{\Gamma(1+5k-5\epsilon)}{\Gamma(1+k-\epsilon)^5}\right|^2,$$ where complex conjugation does not act on $\epsilon$, the coefficient of $\zeta(3)$ term in $Z$ is given by $X^0(z) \overline{ X^0(z)}$ where $$X^0(z)=\sum_k (-z)^k \frac{\Gamma(1+5k)}{\Gamma(1+k)^5}$$ as discussed in eq. 4.6 of this paper ?
$z=0$ is an isolated singularity for $f(z)=\frac{\sin(5\pi z)}{\sin^5(\pi z)}$, in particular a pole of order four with residue zero (since $f$ is an even function). By the Cauchy's integral formula it follows that the integral is just zero.