Let $l$ be a prime number, $n\in \mathbb{Z}$. Is it true that any finitely generated $\mathbb{Z}/l^n\mathbb{Z}$-module has a finite (left) resolution by free finitely generated $\mathbb{Z}/l^n\mathbb{Z}$-modules?
I am not an expert in the field, so the question might not be on the research level. Sorry about that.
No. In fact, this is as far from true as possible: a finitely generated $\mathbb{Z}/l^n\mathbb{Z}$-module has a finite free resolution iff it is free. To see this, note that a finitely generated $\mathbb{Z}/l^n\mathbb{Z}$ is a direct sum of modules of the form $\mathbb{Z}/l^m\mathbb{Z}$ for $m\leq n$ (this follows from the classification of finitely generated $\mathbb{Z}$-modules, for instance). We can construct an infinite periodic free resolution of $\mathbb{Z}/l^m\mathbb{Z}$ like this: $$\dots\stackrel{l^{n-m}}\to\mathbb{Z}/l^n\mathbb{Z}\stackrel{l^m}\to\mathbb{Z}/l^n\mathbb{Z}\stackrel{l^{n-m}}\to\mathbb{Z}/l^n\mathbb{Z}\stackrel{l^m}\to\mathbb{Z}/l^n\mathbb{Z}\to\mathbb{Z}/l^m\mathbb{Z}\to 0.$$
If $0<m<n$ (i.e., if $\mathbb{Z}/l^m\mathbb{Z}$ is not free), then when you Hom this resolution into $\mathbb{Z}/l\mathbb{Z}$ all the maps become $0$, so you find that $\operatorname{Ext}^k(\mathbb{Z}/l^m\mathbb{Z},\mathbb{Z}/l\mathbb{Z})=\mathbb{Z}/l\mathbb{Z}$ for all $k$. Thus $\mathbb{Z}/l^m\mathbb{Z}$ has infinite projective dimension.
