Suppose we have sets $A,B,C$ which are $n$-equidistant.

  Then $n=|A\triangle C|=|(A\triangle B)\triangle (B\triangle C)|$. But $A\triangle B$ and $B\triangle C$ are finite sets of size $n$, and the size of the symmetric difference of two finite sets of equal size is always even.

Why? If $|X|=|Y|=n$, then $|X\triangle Y|=|X\setminus (X\cap Y)|+|Y\setminus (X\cap Y)| = 2(n-|X\cap Y|)$.

Thus $M_n=2$ whenever $n$ is odd.

Suppose we have sets $A,B,C$ which are $n$-equidistant. Then $$n=|A\triangle C|=|(A\triangle B)\triangle (B\triangle C)| = 2(n-|(A\triangle B)\cap (B\triangle C)|),$$ so $n$ is even. Thus $M_n=2$ whenever $n$ is odd.

My answer was incorrect - I didn't notice that the sets could be infinite.

Suppose we have sets $A,B,C$ which are $n$-equidistant.

Then $n=|A\triangle C|=|(A\triangle B)\triangle (B\triangle C)|$. But $A\triangle B$ and $B\triangle C$ are finite sets of size $n$, and the size of the symmetric difference of two finite sets of equal size is always even.

Why? If $|X|=|Y|=n$, then $|X\triangle Y|=|X\setminus (X\cap Y)|+|Y\setminus (X\cap Y)| = 2(n-|X\cap Y|)$.

Thus $M_n=2$ whenever $n$ is odd.