As Henri Cohen remarked, the identity to prove is equivalent to $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{\pi^4}{972}.\tag{1}$$ In turn, this follows readily from the OP's last display (which is a known identity, see e.g. here): $$\left( \sin^{-1}(z)\right)^4=\frac{3}{2}\sum_{k=2}^\infty\frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2 \binom{2k}{k}}, \qquad |z|<1.\tag{2}$$ Let us see how $(2)$ implies $(1)$. We write $k=n+1$ in $(2)$, and observe that $$(n+1)^2\binom{2n+2}{n+1}=2(n+1)(2n+1)\binom{2n}{n}.$$ Therefore, $$\left( \sin^{-1}(z)\right)^4=\frac{3}{4}\sum_{n=1}^\infty\frac{H_n^{(2)}(2z)^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}}, \qquad |z|<1.\tag{2}$$ For $z=1/2$ this gives $(1)$: $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{4}{3}\left( \sin^{-1}\Bigl(\frac{1}{2}\Bigr)\right)^4=\frac{4}{3}\Bigl(\frac{\pi}{6}\Bigr)^4=\frac{\pi^4}{972}.$$

As Henri Cohen remarked, the identity to prove is equivalent to $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{\pi^4}{972}.\tag{1}$$ In turn, this follows readily from the OP's last display (which is a known identity, see e.g. here): $$\left( \sin^{-1}(z)\right)^4=\frac{3}{2}\sum_{k=2}^\infty\frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2 \binom{2k}{k}}, \qquad |z|<1.\tag{2}$$ Let us see how $(2)$ implies $(1)$. We write $k=n+1$ in $(2)$, and observe that $$(n+1)^2\binom{2n+2}{n+1}=2(n+1)(2n+1)\binom{2n}{n}.$$ Therefore, $$\left( \sin^{-1}(z)\right)^4=\frac{3}{4}\sum_{n=1}^\infty\frac{H_n^{(2)}(2z)^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}}, \qquad |z|<1.$$ For $z=1/2$ this gives $(1)$: $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{4}{3}\left( \sin^{-1}\Bigl(\frac{1}{2}\Bigr)\right)^4=\frac{4}{3}\Bigl(\frac{\pi}{6}\Bigr)^4=\frac{\pi^4}{972}.$$

As Henri Cohen remarked, the identity to prove is equivalent to $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{\pi^4}{972}.\tag{1}$$ This, in turn follows readily from the OP's last display (which is a known identity, see e.g. here): $$\left( \sin^{-1}(z)\right)^4=\frac{3}{2}\sum_{k=2}^\infty\frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2 \binom{2k}{k}}, \qquad |z|<1.\tag{2}$$ Let us see how $(2)$ implies $(1)$. We write $k=n+1$ in $(2)$, and observe that $$(n+1)^2\binom{2n+2}{n+1}=2(n+1)(2n+1)\binom{2n}{n}.$$ Therefore, $$\left( \sin^{-1}(z)\right)^4=\frac{3}{4}\sum_{n=1}^\infty\frac{H_n^{(2)}(2z)^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}}, \qquad |z|<1.\tag{2}$$ For $z=1/2$ this gives $(1)$: $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{4}{3}\left( \sin^{-1}(\frac{1}{2})\right)^4=\frac{4}{3}\left(\frac{\pi}{6}\right)^4=\frac{\pi^4}{972}.$$

As Henri Cohen remarked, the identity to prove is equivalent to $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{\pi^4}{972}.\tag{1}$$ In turn, this follows readily from the OP's last display (which is a known identity, see e.g. here): $$\left( \sin^{-1}(z)\right)^4=\frac{3}{2}\sum_{k=2}^\infty\frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2 \binom{2k}{k}}, \qquad |z|<1.\tag{2}$$ Let us see how $(2)$ implies $(1)$. We write $k=n+1$ in $(2)$, and observe that $$(n+1)^2\binom{2n+2}{n+1}=2(n+1)(2n+1)\binom{2n}{n}.$$ Therefore, $$\left( \sin^{-1}(z)\right)^4=\frac{3}{4}\sum_{n=1}^\infty\frac{H_n^{(2)}(2z)^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}}, \qquad |z|<1.\tag{2}$$ For $z=1/2$ this gives $(1)$: $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{4}{3}\left( \sin^{-1}\Bigl(\frac{1}{2}\Bigr)\right)^4=\frac{4}{3}\Bigl(\frac{\pi}{6}\Bigr)^4=\frac{\pi^4}{972}.$$