Let $\newcommand\ZZ{\mathbb Z}R=\ZZ/p^n$, with $n>1$, be your ring and consider the module $M=\ZZ/p$. The obvious map $R\to M$ has kernel the ideal generated by $p$, so the sequence $R\xrightarrow{p}R\to M$ is exact. The kernel of the map $p:R\to R$ is generated by $p^{n-1}\in R$. We thus have an exact sequence of the form $R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\to M$ Now the kernel of $p^{n-1}:R\to R$ is precisely the ideal generated by $p$, so we can extend the exact sequence to $R\xrightarrow{\quad p\quad}R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\to M$ We can continue in this way, and conclude that there is a projective resolution of $M$ as an $R$-module which is of infinite length and with maps $p$ and $p^{n-1}$, repeaing with period $2$: $\cdots \to R\xrightarrow{\quad p\quad}R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\to M$ To compute $Tor$, we drop the $M$ on the right, apply the functor $(-)\otimes_R M$ and compute the homology of the resulting complex. If you do that, you will find that the differentials in the complex are all zero, and all the modules isomorphic to $M$: it follows that $Tor_i(M,M)\cong M$ for all $i\geq0$.