The paper gives (second-to-last page, right-hand column)
$$\sum_{n\ge1}\frac{1}{n^5\binom{2n}{n}}=\frac{9\sqrt{3}\pi}{8}\color{Purple}{L\left(4,\left(-3\atop\circ\right)\right)}+\frac{\pi^2\zeta(3)}{9}-\frac{19\zeta(5)}{3}. \tag{1}$$
You seek a way to write the $L$-function in purple as a linear combination of Hurwitz $\zeta$ functions.
More generally, let $\chi$ be a Dirichlet character of modulus (period) $m$, and define a 'Kronecker' delta
$$\delta_m(k)=\begin{cases}1 & k\equiv0\bmod m \\ 0 & k\not\equiv 0\bmod m\end{cases}. \tag{2}$$
Notice then that $\delta_m(a-b)$ is $1$ if and only if $a\equiv b\bmod m$. We can therefore decompose $\chi$ as
$$\chi(n)=\sum_{k=0}^{m-1} \chi(k) \delta_m(n-k). \tag{3}$$
Furthermore, the Hurwitz zeta function at $a/m\in[0,1)$ decomposes as
$$\begin{array}{c l} \zeta\left(s,\frac{a}{m}\right) & =\sum_{n=1}^\infty\frac{1}{(n+a/m)^s} \\
& =m^s\sum_{n=1}^\infty\frac{1}{(mn+a)^s} \\
& =m^s\sum_{n\ge1} \frac{\delta_m(n-a)}{n^s}.\end{array} \tag{4}$$
Therefore, we have
$$\begin{array}{c l} L(s,\chi) & =\sum_{n\ge1}\frac{\chi(n)}{n^s} \\
& =\sum_{n\ge1}\frac{1}{n^s}\sum_{k=0}^{m-1}\chi(k)\delta_m(n-k) \\
& =\sum_{k=0}^{m-1}\chi(k)\sum_{n\ge1}\frac{\delta_m(n-k)}{n^s} \\
& =\frac{1}{m^s}\sum_{k=0}^{m-1}\chi(k)\zeta\left(s,\frac{k}{m}\right). \end{array} \tag{5}$$
This formula is listed on Wikipedia's Hurwitz $\zeta$ and Dirichlet $L$-function articles. In particular,
$$L\left(4,\left(\frac{-3}{\circ}\right)\right)=\frac{\zeta\left(4,\frac{1}{3}\right)-\zeta\left(4,\frac{2}{3}\right)}{81} \tag{6}$$
because $\left(\frac{-3}{1}\right)=1$ and $\left(\frac{-3}{2}\right)=-1$ (and $\chi(0)=0$ for all Dirichlet characters). Also see here.
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.
Best Answer
Here is my solution:
Let $S$ denote the summation in question. Then by the successive application of integration by parts, we obtain
\begin{align*} S &= 8 \int_{0}^{1} \frac{1}{x} \int_{0}^{x/2} \frac{\arcsin^{2} t}{t} \, dt dx = -8\int_{0}^{1} \frac{\arcsin^{2} (x/2)}{x} \log x \, dx \\ &= 8 \int_{0}^{1} \frac{\arcsin (x/2)}{\sqrt{1 - (x/2)^{2}}} \log^{2} x \, \frac{dx}{2}. \end{align*}
Thus with the substitution $x = 2 \sin\theta$, we have
$$ S = 8 \int_{0}^{\frac{\pi}{6}} \theta \log^{2} (2 \sin\theta) \, d\theta. $$
To evaluate this integral, note that for $0 < \theta < \frac{\pi}{6}$ we have
$$ e^{i\theta} \cdot 2 \sin \theta = i \cdot (1 - e^{2i\theta}). $$
Taking logarithm (with the branch cut $(-\infty, 0]$ as usual) to both sides, it follows that
$$ i\theta + \log (2\sin\theta) = \frac{i\pi}{2} + \log(1 - e^{2i\theta}). $$
Cubing both sides and integrating on $\left( 0, \frac{\pi}{6} \right)$ and taking imaginary parts only,
$$ S = \frac{2}{3} \left( \frac{\pi}{6} \right)^{4} + \frac{8}{3} \Im \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta. \tag{1} $$
Now we focus on the integral in the imaginary part:
$$ I := \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta \tag{2}. $$
Once we evaluate the imaginary part of $I$, the identity $(1)$ immediately gives us the answer. We first make the substitution $z = 1 - e^{2i\theta}$ and $\omega = e^{-i\pi/3}$ to obtain
$$ I = \int_{0}^{\omega} \left( \frac{i\pi}{2} + \log z \right)^{3} \frac{dz}{2i(z-1)}. $$
Here, the path of integration is a circular arc joining from $0$ to $\omega$ centered at $1$ (green-colored path in the figure below).
But since the integrand is analytic for $0 < \Re z < 1$, we may change the path of integration as $z = \omega t$ for $0 \leq t \leq 1$ (blue-colored path in the figure above). This gives
$$ I = \frac{1}{2i} \int_{0}^{1} \left( \frac{i\pi}{6} + \log t \right)^{3} \frac{\omega \, dt}{\omega t - 1}. $$
Plugging $t = e^{-x}$, $I$ reduces to
\begin{align*} I &= \frac{1}{2} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} \frac{\omega e^{-x}}{1 - \omega e^{-x}} \, dx = \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} e^{-nx} \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \left( -\frac{6 i}{n^4}-\frac{\pi }{n^3}+\frac{i \pi ^2}{12 n^2}+\frac{\pi ^3}{216 n} \right). \end{align*}
Taking the imaginary part,
\begin{align*} \Im I &= -3 \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^4} + \frac{\pi}{2} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n^3} + \frac{\pi^2}{24} \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^2} - \frac{\pi^3}{432} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n}. \tag{3} \end{align*}
Note that for $0 < \theta < \pi$, we have
$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \Im \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = - \Im \log(1 - e^{i\theta}) = \frac{\pi-\theta}{2}. $$
Integrating both sides, we obtain
$$ \sum_{n=1}^{\infty} \frac{1-\cos n\theta}{n^{2}} = \frac{\theta (2 \pi -\theta )}{4} \quad \Longrightarrow \quad \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{2}} = \frac{\theta ^2}{4}-\frac{\pi \theta }{2}+\frac{\pi ^2}{6}.$$
Repeating this procedure, we obtain
$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^{3}} = \frac{\theta ^3}{12}-\frac{\pi \theta ^2}{4}+\frac{\pi ^2 \theta }{6}$$
and
$$ \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{4}} = -\frac{\theta ^4}{48}+\frac{\pi \theta ^3}{12}-\frac{\pi ^2 \theta ^2}{12}+\frac{\pi^4}{90}.$$
Plugging $\theta = \frac{\pi}{3}$, we have
\begin{align*} \sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{2}} &= \frac{\pi}{3} \\ \sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{2}} &= \frac{\pi^3}{36} \\ \sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{3}} &= \frac{5 \pi^3}{162} \\ \sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{4}} &= \frac{91 \pi ^4}{19440} \end{align*}
Plugging these to $(3)$, we have
$$ \Im I = \frac{23 \pi^4}{12960} \quad \Longrightarrow \quad S = \frac{17 \pi^4}{3240} = \frac{17}{36}\zeta(4) $$
as desired.