On binomial sums $\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$ and log sine integrals
Solution 1:
Persistence pays off! Given the log sine integral,
$$\rm{Ls}_n\Big(\frac{\pi}3\Big) = \int_0^{\pi/3}\ln^{n-1}\big(2\sin\tfrac{\theta}{2}\big)\,d\theta$$
The case $\rm{Ls}_7\big(\frac{\pi}3\big)$ was elusive, but $\rm{Ls}_\color{red}8\big(\frac{\pi}3\big)$ was found. Hence,
$$\frac{2^{10}\cdot9\pi}{7!} \int_0^{\pi/3}\ln^7\big(2\sin\tfrac{x}{2}\big)\,dx+6^3 \sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n}\\=-13921\zeta(9)-6^4\zeta(2)\zeta(7)-6087\zeta(3)\zeta(6)-4428\zeta(4)\zeta(5)-192\zeta^3(3) $$
though I don't know why the lower level is more elusive.