Suppose that $A,B$ are two positive $n \times n$ complex matrices. Consider the operator $$ A + i B $$ which is not necessarily normal. However, if $\lbrack A, B \rbrack = 0$ then $A + i B$ is normal.
When this is not the case, $A+ iB$ nevertheles has a Jordan normal form. There exists an invertible matrix $T$ such that $$ A + i B = T D T^{-1} $$ for a diagonal matrix.
The condition number is defined by $$ \kappa(A+iB) = \vert \vert{T}\vert \vert \vert \vert T^{-1} \vert \vert. $$ Whenever $A+iB$ is normal it holds that $\kappa(A+iB) =1$.
Is there a more general relation between $\kappa(A+iB)$ and $\vert \vert \lbrack A, B \rbrack \vert \vert$ capturing the degree of non-normality?
This is extremely unlikely. Take $A$ as any $n$ by $n$ matrix and let $B$ be the $n$ by $n$ zero matrix. Then $AB=BA=0$, so $\|[A,B]\| = 0$ for any matrix norm. From this zero we can deduce nothing about the conditioning of $A$. Conversely, if $A = B = \text{diag}(1,t)$ are 2-by-2 diagonal matrices, then the singular values of $T = A+iB$ are simply $\sqrt{2}$ and $\sqrt{1+t^2}$ and the 2-norm condition number can assume any value in $[\sqrt{2},\infty)$, while $[A,B]=0$ for any $t$.