For finite matrices, your norm is the spectral radius of $A$. Indeed, for each matrix $A$ you can construct a Jordan-like form $A=VJV^{-1}$ in which the strictly upper diagonal part contains $\varepsilon$ instead of 1; then define a norm $\|M\|:=\|V^{-1}MV\|_2$ (where $\|K\|_2$ is the operator norm of $K$), and then $\|A\| = \rho(A) + O(\varepsilon)$. On the other hand, each norm is greater than the spectral radius.

For finite matrices, your 'norm' is the spectral radius of $A$. Indeed, one can construct for each matrix $A$ a matrix norm induced by a vector norm such that $\|A\| \leq \rho(A) + \varepsilon$ for each $\varepsilon>0$. (And, on the other hand, $\|A\|\geq \rho(A)$ for each norm induced by a vector norm).

1. for each matrix $A$ you can construct a Jordan-like form $A=VJV^{-1}$ in which the strictly upper diagonal part contains $\varepsilon$'s instead of 1 (you can get it by taking the Jordan form and conjugating by $\operatorname{diag}(1,\varepsilon,\varepsilon^2,\dots,\varepsilon^{n-1})$).
2. then define the norm $\|M\|:=\|V^{-1}MV\|_2$ (where $\|K\|_2$ is the operator norm of $K$), which is induced by the vector norm $\|x\| = \|V^{-1}x\|$. Then $\|A\| = \rho(A) + O(\varepsilon)$. On the other hand, each norm is greater than the spectral radius.

This is a 'standard' proof trick that I learned in my courses.