I would extend Jason's comment to say that this condition trivially holds in any Artinian ring since in those rings every intersection of ideals is a finite intersection. It seems that indeed this is the only case. 


Let $R$ be an arbitrary noetherian ring and $\mathfrak m$ an arbitrary maximal ideal in $R$. If $R$ is not an integral domain, let $\mathfrak p\subset R$ be a minimal prime ideal contained in $\mathfrak m$ and let $A:=R/\mathfrak p$.
Clearly $A$ is an integral domain and $\overline{\mathfrak m}:=\mathfrak m/\mathfrak p$ is a maximal ideal in $A$. By the Krull intersection theorem $\cap_n \overline{\mathfrak m}^n=0$. 

It follows that then $\cap_n\mathfrak m^n\subseteq \mathfrak p$ in $R$. If $R$ satisfied the desired condition, then this would imply that  $\mathfrak m^n\subseteq \mathfrak p$ for some fixed $n\in \mathbb N$. But then $\mathfrak m= \mathfrak p$ is both maximal and minimal. This is true for every maximal ideal, so $\dim R=0$ and hence $R$ is an Artinian. 

I leave it for you to decide about the non-noetherian case.