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.

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.

EDIT 2: Sorry, I made a stupid mistake in the struck-out sentence above. Of course, if $ M \colon X \to X^\ast $, then we again have $ M^\ast \colon X \to X^\ast $ -- not $X^\ast \to X$ as I had written above. Ultimately, you may have to resort to the fact that $\ell^p$ is isomorphic to $\ell^2$ (since $n$ is finite), so one can map between $X$ and $X^\ast$ -- but this has gotten sufficiently far from my original answer that I'll just stop at that.

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.

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.