What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions?

We have the following evaluations:

$$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\&\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = \,?\\ \end{aligned}$$

The paper Borwein and Bradley's Apery-like Formula for $\zeta(4n+3)$ gives the 3rd and 5th in terms of the Dirichlet L-functions, but does anyone know how to evaluate the 5th one in terms of the Hurwitz zeta function $\zeta(s,a)$?

Postscript: (A few hours later)

After Anon gave his answer, I did a little more sleuthing and found the case p = 7 in the Mathworld article on central binomial coefficients (which also had p = 5). The paper I cited was a bit old (1999) and the authors weren’t aware it was already found a year earlier by Plouffe. Hence,

$$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n^5 \, \binom{2n}n} = -\frac{19}{3}\zeta(5)+\frac{2}{3}\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot3^2}\left(\zeta(4,\tfrac{1}{3})-\zeta(4, \tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^7 \, \binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6, \tfrac{2}{3}) \right)\\ \end{aligned}$$

With this “pattern”, I used an integer relations algorithm to try to find p = 9, 11, 13. No luck so far.

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.