Closed form for a zeta series :$\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$ It is not that diffcult to derive
\begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{k2^k}=&-\frac{\gamma}{2}+\ln\left(\frac{2}{\sqrt{\pi}}\right)\tag{1}\\
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+1)2^{k+1}}=&-\frac{4+\gamma}{8}+\ln\left(A^{3/2}2^{5/24}\right)\tag{2}
\end{align}
Hence, I would like to know if there exists a closed form in terms of known mathematical constants for the following series
$$\mathscr{S}=\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$$
As $(1)$ and $(2)$ follow immediately from the definitions of $\Gamma(z)$ and $G(z+1)$ respectively, my guess is that the evaluation of $\mathscr{S}$ involves the function $\Gamma_3(z)$. Unfortunately, I know almost nothing about higher order multiple gamma functions, and I would really appreciate it if someone can enlighten me on this matter and provide a viable solution to the series above. Thank you.

This is what I have managed to get so far.
Begin with the sum
\begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{k+1}z^{k+1}
=&\sum^\infty_{k=3}\sum^\infty_{m=1}\frac{(-1)^k}{k}\frac{z^{k}}{m^{k-1}}\\
=&\sum^\infty_{m=1}\left\{-m\ln\left(1+\frac{z}{m}\right)-\frac{z^2}{2m}+z\right\}\\
\end{align}
Compare this with $\ln{G(z+1)}$.
$$\ln{G(z+1)}=-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{m=1}\left\{m\ln\left(1+\frac{z}{m}\right)+\frac{z^2}{2m}-z\right\}$$
It follows that
$$\sum^\infty_{k=1}\frac{(-1)^{k-1}\zeta(k)}{k+1}z^{k+1}=-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}-\ln{G(z+1)}$$
Integrate from $0$ to $z$ to get
\begin{align}
&-\frac{z^2}{2}+\frac{z^2}{2}\ln(2\pi)-\frac{z^3}{2}-\frac{\gamma z^3}{2}-z\ln{G(z+1)}-\sum^\infty_{k=1}\frac{(-1)^{k-1}\zeta(k)}{(k+2)}z^{k+2}\\
=&\sum^\infty_{k=1}\left\{-k(k+z)\ln\left(\frac{k+z}{k}\right)+kz+\frac{z^2}{2}-\frac{z^3}{6k}\right\}
\end{align}
After letting $z=\frac12$, I have no idea how to proceed further as when I take the exponential of the partial sum, the portion with the $\ln$ term doesn't seem to telescope.

As mentioned by Claude Leibovici, you have

$$
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}=-\frac{1}{8}-\frac{\gamma}{24}+\frac{\ln 2}{12}+\frac{\ln A}{2}-\frac{7\zeta(3)}{16 \pi ^2}. \tag1
$$

Here is a hint.
From  the classic identity verified by the digamma function $\displaystyle \psi:=\Gamma'/\Gamma$, wich may be obtained from the Euler product giving $\Gamma(x+1)$, you have
$$ \psi(x+1) = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k(k+x)}\quad x\neq 0,-1,-2,-3,\dots$$
then you easily obtain, for $|x|<1$,
$$ \begin{align}
\psi(x+1) & = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k^2}\frac{1}{1+\dfrac{x}{k}} \\
&= -\gamma + \sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=0}^{\infty}\frac{(-1)^n}{k^n}x^{n+1} \\
&= -\gamma - \sum_{n=0}^{\infty}(-1)^{n-1} \zeta(n+2){x^{n+1}} \\
&= -\gamma - \sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k-1}} \\
\end{align}
$$
and
$$-\gamma x^2 - x^2\psi(x+1) = \sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k+1}}. \tag2$$
Using $(2)$ gives
$$
\begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}&=\sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k)\int_0^{1/2}\!\!x^{k+1}dx\\
&= \int_0^{1/2}\!\sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k+1}}\:dx \\
&= -\gamma \int_0^{1/2}\!x^2 dx - \int_0^{1/2}\! x^2\psi(x+1)\:dx \\
&=-\frac{\gamma}{24} - \int_0^{1/2}\! x^2\psi(x+1)\:dx,
\end{align}
$$ then integrating by parts twice leads to 
$$ \begin{align}
\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}} & = -\frac{\gamma}{24} - \frac14\log \Gamma\left(\frac32\right)+2\int_0^{1/2}\! x\log \Gamma(x+1)\:dx\\
&=  -\frac{\gamma}{24} - \frac18\log \pi- \frac14\ln2+\zeta'\left(-1,\frac32\right)+2\int_0^{1/2}\! \zeta'(-1,x+1)\:dx \\
&= -\frac{\gamma}{24}-\frac{1}{8}+\frac{\ln 2}{12}+\frac{\ln A}{2}-\frac{7\zeta(3)}{16 \pi ^2},
\end{align}
$$
where we have used the identity (25.11.34)  and special values of $\zeta'(s,a)$.
