$$ \pi = \sum_{n=0}^{\infty} \frac{1}{16^n}\left(\frac{4}{8n+1} -\frac{2}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}\right)$$
I guess that using the Leibniz series could be somehow the way of the proof
$$\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1}=1-\frac{1}{3}+\frac{1}{5}-...$$ is it the right way of thinking?
Use $\int_0^1x^{n-1}dx=\frac1n$ to write your sum as$$\begin{align}&\int_0^1\sum_{n\ge0}\left(\frac{x^8}{16}\right)^n(4-2x^3-x^4-x^5)dx\\&\stackrel{\ast}{=}\int_0^1\frac{16(1-x)}{(2-x^2)(2-2x+x^2)}dx\\&=\int_0^1\left(-\frac{4x}{2-x^2}+\frac{8-4x}{2-2x+x^2}\right)dx\\&=[2\ln(2-x^2)-2\ln(2-2x+x^2)+4\arctan(x-1)]_0^1\\&=\pi,\end{align}$$where $\stackrel{\ast}{=}$ repeatedly uses the difference of two squares to take common factors out of the rational integrand (note we've summed a geometric series). That's the hard part, which you should work through yourself.