It might be more useful to pose the problem as follows. Let $X = (\mathbb{R}^n, \| \cdot \| _p )$ and $X^\ast = (\mathbb{R}^n, \| \cdot \| _{p^\ast})$, where $p^\ast$ is the conjugate exponent to $p$. Rather than considering $M$ as a map from $ X \to X $, it may be more useful to treat it as $M \colon X \to X^\ast $. (Of course, when $p=p^{*}=2$, these are the same.) In that case, one can make sense of the compositions $M^\ast M \colon X \to X$ and $M M^\ast \colon X ^\ast \to X^\ast $, and take the singular values as the square root of the eigenvalues of these maps.
EDIT: This is equivalent to looking at $\| Mx \| _{p^\ast} / \| x \|_p $ instead of $\| Mx \| _p / \| x \|_p $, so it ties into the work that Suvrit mentioned in his response.