Let $R$ denote the circumradius of triangle $ABC$. As $ABC$ is an acute triangle, it isn't hard to check that for any point $P$, we have either $AP \geq R, BP \geq R$, or $CP \geq R$. Also, note that if we choose $P=O$ (the circumcenter) then $\left(AP^{n}+BP^{n}+CP^{n}\right)=3 \cdot R^{n}$. Therefore, we have the inequality $R \leq \min _{P \in \mathbb{R}^{2}}\left(AP^{n}+BP^{n}+CP^{n}\right)^{\frac{1}{n}} \leq\left(3 R^{n}\right)^{\frac{1}{n}}=R \cdot 3^{\frac{1}{n}}$. Taking $n \rightarrow \infty$ yields $R \leq \lim _{n \rightarrow \infty} M_{n} \leq R$ (as $\lim _{n \rightarrow \infty} 3^{\frac{1}{n}}=1$ ), so the answer is $R=\frac{65}{8}$.