Consider the sums
$$S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4\over \pi\right)^{2n}}\cdot\sum_{k=0}^{\infty}{(-1)^{k(k+1)\over 2}\over (2k+1)^{2n}}$$
We have $S(1)=1$, $S(2)=11$, $S(3)=361$, $S(4)=24611$.
I cannot spot any pattern within this sequence.
How can we work out a closed form for $S(n)$?
We may notice that $$ T(n)=\sum_{k\geq 0}\frac{(-1)^{k(k+1)/2}}{(2k+1)^{2n}} = \sum_{m\geq 1}\frac{\chi(m)}{m^{2n}} = L(\chi,2n)\tag{1}$$ is a Dirichlet $L$-function associated with the multiplicative function $\chi(m)$, that equals $0$ if $m$ is even, $1$ if $m\equiv \pm 1\pmod{8}$ and $-1$ if $m\equiv \pm 3\pmod{8}$. In particular $$ T(n) = \prod_{p>2}\left(1-\frac{\left(\frac{2}{p}\right)}{p^{2n}}\right)^{-1}\tag{2} $$ where $\left(\frac{2}{p}\right)$ is Legendre's symbol. From Hazem Orabi's integral representation $$ S(n) = \frac{1}{\sqrt{2}(2\pi)^{2n}}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{x/8}-e^{3x/8}-e^{5x/8}+e^{7x/8}\right)\,dx \tag{3} $$ we also have: $$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\int_{1}^{+\infty}\log(x)^{2n-1}\frac{x^2-1}{1+x^4}\,dx \tag{4}$$ or $$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\left.\frac{d^{2n-1}}{d\alpha^{2n-1}}\int_{1}^{+\infty}\frac{x^{2+\alpha}-x^{\alpha}}{1+x^4}\,dx\, \right|_{\alpha=0^+}\tag{5}$$ so $S(n)$ depends on $\psi^{(2n-1)}(z)$ (the $(2n-1)$-th derivative of the digamma function, i.e. the $2n$-th derivative of $\log\Gamma$) evaluated at $z=\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}$. By the reflection formulas for the $\psi^{(2n-1)}(z)$ function, everything boils down to the derivatives of the function $\cot(\pi z)$ at $z=\frac{1}{8}$ and $z=\frac{3}{8}$. This proves G.H.Hardy's claim:
$$ S(n) = (2n-1)!\cdot [z^{2n-1}]\frac{\sin(z)}{\cos(2z)}.\tag{6} $$
In particular, $S(n)$ can be expressed in terms of Euler's polynomials $E_n(z)$ evaluated at $z=\frac{1}{4}$ and $z=\frac{3}{4}$.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{S}\pars{n} & \equiv {\pars{2n - 1}! \over \root{2}}\,\pars{4 \over \pi}^{2n}\sum_{k = 0}^{\infty} {\pars{-1}^{k\pars{k + 1}/2} \over \pars{2k + 1}^{2n}} \\[5mm] & = \pars{2n - 1}!\,{2^{4n - 1/2} \over \pi^{2n}}\bracks{% \sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{4k + 1}^{2n}} + \sum_{k = 0}^{\infty}{\pars{-1}^{k + 1} \over \pars{4k + 3}^{2n}}} \\[5mm] & = {\root{2} \over 2\pi^{2n}}\bracks{% \pars{2n - 1}!\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + 1/4}^{2n}} - \pars{2n - 1}!\,\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + 3/4}^{2n}}} \label{1}\tag{1} \end{align}
However,
\begin{align} &\left.\pars{2n - 1}!\,\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + a}^{2n}}\right\vert_{\ a\ >\ 0} = \pars{2n - 1}!\,\sum_{k = 0}^{\infty}\pars{-1}^{k}\bracks{% {1 \over \Gamma\pars{2n}} \int_{0}^{\infty}t^{2n - 1}\expo{-\pars{k + a}t}\,\dd t} \\[5mm] = &\ \int_{0}^{\infty}t^{2n - 1}\expo{-at}\sum_{k = 0}^{\infty} \pars{-\expo{-t}}^{k}\,\dd t = \int_{0}^{\infty}t^{2n - 1}\expo{-at}\,{1 \over 1 + \expo{-t}}\,\dd t \label{2}\tag{2} \\[5mm] = &\ \bbx{\ds{4^{-n}\,\Gamma\pars{2n}\bracks{% \zeta\pars{2n,{a \over 2}} - \zeta\pars{2n,{a + 1 \over 2}}}}} \end{align}
The last integral, in \eqref{2}, is straightforward evaluated by expanding $\ds{1 \over 1 + \expo{-t}}$ in powers of $\ds{\expo{-t}}$. Expression \eqref{1} is reduced to:
\begin{align} \mrm{S}\pars{n} & \equiv {\pars{2n - 1}! \over \root{2}}\,\pars{4 \over \pi}^{2n}\sum_{k = 0}^{\infty} {\pars{-1}^{k\pars{k + 1}/2} \over \pars{2k + 1}^{2n}} \\[5mm] & = \bbx{\ds{{\root{2} \over 2}\,{\Gamma\pars{2n} \over \pars{2\pi}^{2n}}\bracks{% \zeta\pars{2n,{1 \over 8}} - \zeta\pars{2n,{5 \over 8}} - \zeta\pars{2n,{3 \over 8}} + \zeta\pars{2n,{7 \over 8}}}}} \end{align}
The above series is the expansion of the function Sin[x]/Cos[2x]
$$ \begin{align}
\sum_{k=0}^{\infty}\frac{(-1)^{\frac{\large k(k+1)}{2}}}{(2k+1)^{2n}} &= \sum_{k=0}^{\infty}\left[\small\frac{(-1)^{\frac{\large (4k+0)(4k+1)}{2}}}{(2(4k+0)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+1)(4k+2)}{2}}}{(2(4k+1)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+2)(4k+3)}{2}}}{(2(4k+2)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+3)(4k+4)}{2}}}{(2(4k+3)+1)^{2n}}\normalsize\right] \\[3mm]
&= \sum_{k=0}^{\infty}\left[\frac{1}{(8k+1)^{2n}}\color{red}{-}\frac{1}{(8k+3)^{2n}}\color{red}{-}\frac{1}{(8k+5)^{2n}}\color{red}{+}\frac{1}{(8k+7)^{2n}}\right] \\[3mm]
&= \frac{1}{8^{2n}}\left[\zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right)\right]
\end{align} $$
Hence,
$$
S(n)=\color{red}{\frac{(2n-1)!}{\sqrt{2}\,(2\pi)^{2n}}\,\left[\zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right)\right]}
$$
It is also important to mention that it seems their are some relation with:
$$ \zeta(2n)= \frac{1}{(2n-1)!}\,\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\,dx = \frac{\left|{B}_{2n}\right|}{2}\,\frac{(2\pi)^{2n}}{(2n)!} $$
One would notice $\,\left(\frac18+\frac78\right)-\left(\frac38+\frac58\right)=\frac88-\frac88=0\,$:
$$ \small \zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right) = \frac{1}{(2n-1)!}\,\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{\frac78x}-e^{\frac58x}-e^{\frac38x}+e^{\frac18x}\right)dx $$