We can decompose Z mod p^n as a vector space over Z mod p by writing numbers base p, eg, mod 3, 52= 27 + 2*9 + 2*3 + 1, and if Rm is {0,p^m,2p^m,....(p-1)p^m}, then we have for a in Rm, b in Rk, that ab is in R(m+k), which seems very similar to the definition of a graded ring.
This extra structure gives rise to a simple proof of the group of units being cyclic (when p is an odd prime), so seems like helpful additional information to know about.
However since each subspace in this decomposition is not closed under addition, it seems like this isn't a graded ring, in which case, what is it?
What I think you are getting at is that $\newcommand{\Z}{\Bbb Z}\Z/p^n\Z$ is a filtered ring and that filtered rings have associated graded rings. A filtration of a commutative ring $R$ is a descending sequence of ideals $I_0\supseteq I_1\supseteq I_2\supseteq\cdots$ such that $I_0=R$ and $I_m I_n\subseteq I_{m+n}$. The associated graded ring is $S=\bigoplus_{n=0}^\infty I_n/I_{n+1}$. It has a multiplication which maps $(I_m/I_{m+1})\times (I_n/I_{n+1})$ to $I_{m+n}/I_{m+n+1}$.
In your case, $I_k=p^k\Z/p^n\Z$ and each $I_k/I_{k+1}$ is a vector space over $I_0/I_1\cong \Z/p\Z$, of dimension $1$ for $k <n$.