The fourier coefficients for nonzero $n$ are
$\begin{eqnarray} \hat f(n) &=& \frac{1}{2 \pi} \int_0^{2 \pi} (\pi-t) e^{-i n t} dt \\ &=& \frac{1}{2} \int_0^{2 \pi} e^{-i n t} dt - \frac{1}{2 \pi} \int_0^{2 \pi} t e^{-i n t} dt \\ &=& - \frac{1}{2 \pi} \left(\left[t \frac{e^{- i n t}}{- i n}\right]_0^{2 \pi} - \int_0^{2 \pi} \frac{e^{- i n t}}{- i n} dt \right) \\ &=& \frac{1}{-in} \\ \end{eqnarray}$
and $\hat f(0) = 0$.
Thus applying Parseval's identity we have
$\sum_{n=-\infty,n\neq 0}^\infty \frac{1}{n^2} = \frac{1}{2\pi}\int_{-\pi}^\pi (\pi - t)^2 \, dt = \frac{1}{2\pi}\int_{0}^{2\pi} t^2 \, dt = \frac{\pi^{2}}{3}$