The only possible value of $n$ is $3$. It is easy to see taking $a_i = m^{i-1}$ for large enough $m$. For $n=3$ the inequality is \[\frac{a_1 + a_2 + a_3}{\sqrt[3]{a_1 a_2 a_3}} \leq M \left( \frac{a_2}{a_1}+\frac{a_3}{a_2}+\frac{a_1}{a_3}\right).\] Let $x=a_1/\sqrt[3]{a_1 a_2 a_3}$, $y=a_2/\sqrt[3]{a_1 a_2 a_3}$, $z=a_3/\sqrt[3]{a_1 a_2 a_3}$. Then $xyz=1$ and \[x+y+z \leq M \left( \frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right).\] It is clear that $M\geq 1$ (take $x=y=z=1$). Prove that $\min M=1$. It is enough to show that for positive $x, y, z$ with $xyz=1$, \[x+y+z \leq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}.\] or \[y\left(\frac{1}{x}-1\right) + z\left(\frac{1}{y}-1\right) + x\left(\frac{1}{z}-1\right) \geq 0.\] Now if we take $a=1/x, b=1/y, c=1/z$, then $abc=1$ and the required inequality takes form \[\frac{a-1}{b} + \frac{b-1}{c} + \frac{c-1}{a} \geq 0\] which is a known beast.

I believe you are right! Also I was wrong; I meant that the difficulty was finding $n$ (i.e. proving $n=3$). Many of us conjectured that $n=3$ and found the minimal value of $M_n$, but failed to prove that $n=3$. (Actually I was one of them)