This is only meant to give some ideas on how to tackle the series via integrals. I hope that someone can further build up on this and finish the solution at some point. To get a first impression, I'd recommend to peek at $(1)$, $(2)$ and $(3)$ directly.
We will start by converting the series into a quadruple integral by utilizing the following two identities:
$$\frac{\arcsin \left( x\right)}{\sqrt{1- x^2}} =\frac12 \sum_{n=1}^\infty \frac{4^n x^{2n-1}}{n\binom{2n}{n}};\quad \int_0^\frac{\pi}{2} \sin^{2n-1}xdx= \frac12\frac{4^n}{n\binom{2n}{n}}$$
$$\Rightarrow \frac{1}{16}\sum_{n=1}^\infty \left(\frac{4^n}{n\binom{2n}{n}}\right)^4 t^n=\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{\arcsin(t\sin x\sin y\sin z)}{\sqrt{1-t^2\sin^2 x\sin^2 y \sin^2 z }}dxdydz$$
$$ \Rightarrow \mathcal S=\sum_{n=1}^\infty \frac{(4n-1) 256^n}{(2n-1)n^5 \binom{2n}{n}}=\small{32\int_0^1\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2} \frac{(1+t)\arcsin(t\sin x\sin y\sin z)}{t\sqrt{1-t^2\sin^2 x\sin^2 y \sin^2 z }}dxdydzdt}$$
Next, we will substitute $\sin x\to x$, $\sin y\to y$ and $\sin z\to z$ followed by $xyz\to x$ to arrive at:
$$\mathcal S=32\int_0^1 \int_0^1 \int_0^1 \int_0^{yz} \frac{(1+t)\arcsin(tx)}{t\sqrt{1-t^2 x^2}} \frac{dxdydzdt}{\sqrt{y^2 z^2-x^2}\sqrt{1-y^2}\sqrt{1-z^2}}$$
$$=32\int_0^1 \frac{(1+t)\arcsin(tx)}{t\sqrt{1-t^2 x^2}} \underbrace{\int_{x}^1 \int_{x/z}^1 \frac{dydzdxdt}{\sqrt{y^2z^2-x^2}\sqrt{1-y^2}\sqrt{1-z^2}}}_{\mathcal J(x)}$$
$$\small =\boxed{16\int_0^1 \int_0^1 \left(K^2\left(\sqrt{\frac{1+\sqrt{1-x^2}}{2}}\right)-K^2\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)\right) \frac{(1+t)\arcsin(tx)}{t\sqrt{1-t^2 x^2}} dxdt} \tag 1$$
Where $K(k)$ is the Complete Elliptic Integral of the First Kind with elliptic modulus $k$. Note that you need to put $K(k)=K(k^2)$ if you want to verify numerically in the Wolfram language.
Note that the $\mathcal J(x)$ integral from above was calculated as follows:
$$\mathcal J(x)=\int_{x}^1 \frac{1}{z\sqrt{1-z^2}}\int_{x/z}^1 \frac{dydz}{\sqrt{y^2-x^2/z^2}\sqrt{1-y^2}}$$
$$\overset{y\to \sqrt{1-(1-x^2/z^2)y^2}}=\int_x^1 \frac{1}{z\sqrt{1-z^2}}\int_0^1 \frac{dydz}{\sqrt{1-y^2}\sqrt{1-(1-x^2/z^2)y^2}}$$
$$=\int_x^1 \frac{K\left(\sqrt{1-\frac{x^2}{z^2}}\right)}{z\sqrt{1-z^2}}dz\overset{\sqrt{1-\frac{x^2}{z^2}}\to z}=\int_0^\sqrt{1-x^2} \frac{zK(z)}{\sqrt{1-z^2}\sqrt{1-x^2-z^2}}dz$$
$$=\Re\int_0^1 \frac{zK(z)}{\sqrt{1-z^2}\sqrt{1-x^2-z^2}}dz \overset{\small \bigstar}= \Re \frac{1}{1+\sqrt{1-x^2}} K^2 \small{\left(\sqrt{\frac{2}{1+\sqrt{1-x^2}}}\right)}$$
$$\small \overset{\small \bigstar \bigstar}=\frac12\left(K^2\left(\sqrt{\frac{1+\sqrt{1-x^2}}{2}}\right)-K^2\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)\right)$$
The $\small \bigstar$ result was derived by Yajun Zhou in this article, see page 19 under eq. $(47)$. Whereas the $\small \bigstar \bigstar$ identity follows by squaring $K(1/k)=k(K(k)-iK'(k))$ and taking the real part.
Now, the most intuitively way to move forward I think would be to compute the integral with respect to $t$, which can be done with the substitution $tx\to t$ followed by $t\to \frac{2t}{1+t^2}$, to obtain:
$$\int_0^1 \frac{(1+t)\arcsin(xt)}{t\sqrt{1-x^2t^2}}dt=2\operatorname{Ti}_2\left(\frac{x}{1+\sqrt{1-x^2}}\right)+\frac{\arcsin^2 x}{2x}$$
Where $\operatorname{Ti}_2(x)$ is the inverse tangent integral. This gives yet another integral representation:
$$\small \mathcal S=32\int_0^1 \left(K^2\left(\sqrt{\frac{1+\sqrt{1-x^2}}{2}}\right)-K^2\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)\right)\left(\operatorname{Ti}_2\left(\frac{x}{1+\sqrt{1-x^2}}\right)+\frac{\arcsin^2 x}{4x}\right)dx$$
$$\overset{x\to \sin x}=\boxed{32\int_0^\frac{\pi}{2} \left(K^2\left(\cos \frac{x}{2}\right)-K^2\left(\sin \frac{x}{2}\right)\right)\left(\operatorname{Ti}_2\left(\tan \frac{x}{2}\right)+\frac{x^2}{4\sin x}\right)\cos x \, dx} \tag2$$
I was quite optimistic about the above integral, especially since it allows to utilize various series such as $\operatorname{Ti}_2(\tan x)-x\ln(\tan x)=\sum\limits_{n=0}^\infty \frac{\sin((4n+2)x)}{(2n+1)^2}$, but felt short in the end.
An even more promising idea is based on another article due to Yajun Zhou, which proves that:
$$\zeta(5)=-\frac{\pi^4}{372}\int_{-1}^1 x\left(P_{-1/2}(x)\right)^4 dx; \quad \zeta(3) = -\frac{\pi^4}{168}\int_{-1}^1 x\left(P_{-1/4}(x)\right)^4 dx$$
Where $P_n(x)$ is the Legendre function, defined via the Mehler-Dirichlet formula. Now, by rewriting $P_{-1/2}(\cos x)=\frac{2}{\pi} K\left(\sin \frac{x}{2}\right)$ and $P_{-1/4}(\cos x)= \frac{2}{\pi \sqrt{1+\sin \frac{x}{2}}}K\left(\sqrt{\frac{2\sin \frac{x}{2}}{1+\sin \frac{x}{2}}}\right)$ followed by substituting $\frac{1-x}{2}\to x$ we arrive at the following result, which surprisingly is identical with the result of our original series:
$$\boxed{32\int_0^1 (2x-1) K^4(\sqrt x)dx-32\zeta(2)\int_0^1 \frac{2x-1}{{(1+\sqrt x)^2}} \small{K^4\left(\sqrt{\frac{2 \sqrt x}{1+\sqrt x}}\right)}dx=372\zeta(5)-28 \pi^2\zeta(3)}$$
As such, it is reasonable to think that we can actually go ahead and connect our previous double integral seen in $(1)$ with the above result. Namely, to arrive at squared elliptic integrals in $(1)$ we integrated twice after essentially substituting $xy \to x$ and $xz\to x$, similarly we can expect that if we further substitute $xt \to x$ we will obtain third powers, and so on for fourth powers.
Anyway, continuing from $(1)$ we have:
$$\small \mathcal S =16\int_0^1 \int_0^1 \left(K^2\left(\sqrt{\frac{1+\sqrt{1-x^2}}{2}}\right)-K^2\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)\right) \frac{(1+t)\arcsin(tx)}{t\sqrt{1-t^2 x^2}} dxdt$$
$$\small \overset{xt \to x}=16\int_0^1 \frac{\arcsin x}{\sqrt{1-x^2}}\int_x^1 \left(\frac{1}{t}+\frac{1}{t^2}\right) \left(K^2\left(\sqrt{\frac{1+\sqrt{1-x^2/t^2}}{2}}\right)-K^2\left(\sqrt{\frac{1-\sqrt{1-x^2/t^2}}{2}}\right)\right) dtdx$$
$$\small \overset{\sqrt{1-x^2/t^2}\to t}=16\int_0^1 \frac{\arcsin x}{\sqrt{1-x^2}} \int_0^\sqrt{1-x^2} \left(\frac{t}{1-t^2}+\frac{t}{x\sqrt{1-t^2}}\right)\left(K^2\left(\sqrt{\frac{1+t}{2}}\right)-K^2\left(\sqrt{\frac{1-t}{2}}\right)\right) dtdx$$
$$\overset{\large {x\to \sin x \atop \large t\to\cos t}}=\boxed{16\int_0^\frac{\pi}{2} x \int_x^\frac{\pi}{2} \left(\frac{\cos t}{\sin t}+\frac{\cos t}{\sin \color{red}{x}}\right)\left(K^2\left(\cos \frac{t}{2}\right)-K^2\left(\sin \frac{t}{2}\right)\right) dtdx} \tag 3$$
Unfortunately, here I hit another dead end as I'm not sure how to further calculate the above $t$ integral, which I suspect that can be expressed as a sum of cubic elliptic integrals.
