Calculate $\sum_{n=0}^\infty \frac1{(4n^2 - 1)^2}$

Let's continue with the approach you took. We have $$\begin{align}\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} &=\frac{1}{4}\sum_{n=0}^\infty \bigg(\frac{1}{2n-1}-\frac{1}{2n+1}\bigg)^2\\ &=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}-\frac{2}{(2n-1)(2n+1)} \end{align}$$

Now, using the fact that $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}$$ we have $$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+2}{8}-\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n-1)(2n+1)}$$ Now we may use telescoping to evaluate the latter series as $$\frac{1}{2}\sum_{n=0}^\infty \frac{1}{(2n-1)(2n+1)}=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}=-\frac{1}{4}$$ so that we have $$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+4}{8}=\frac{1}{2}+\frac{\pi^2}{16}$$

Note that$$(\forall n\in\mathbb{N}):\frac1{(4n^2-1)^2}=\frac1{4(2n+1)}-\frac1{4 (2n-1)}+\frac1{4(2n+1)^2}+\frac1{4(2n-1)^2}.$$It is clear that the series$$\sum_{n=0}^\infty\frac1{4(2n+1)}-\frac1{4(2n-1)}$$is a telescoping series, whose sum is $\frac14$. On the other hand\begin{align}\sum_{n=0}^\infty\frac{1}{4 (2 n+1)^2}+\frac{1}{4(2 n-1)^2}&=\frac14+2\sum_{n=0}^\infty\frac{1}{4 (2 n+1)^2}\\&=\frac14+\frac12\sum_{n=0}^\infty\frac1{(2n+1)^2}.\end{align}But, since$$\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8,$$we have that the sum of your series is$$\frac14+\frac14+\frac{\pi^2}{16}=\frac12+\frac{\pi^2}{16}.$$

Since $$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1} $$ by Parseval's theorem we have $$ \pi=\int_{-\pi}^{\pi}\left|\sin x\right|^2\,dx =2\pi\cdot\frac{4}{\pi^2}+\frac{16}{\pi}\sum_{n\geq 1}\frac{1}{(4n^2-1)^2}$$ hence by rearranging: $$ \sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \color{red}{\frac{\pi^2}{16}-\frac{1}{2}}.$$