Is there a geometric meaning of the Frobenius norm?

To give intuition, lets look at a particularly nice class of matrices. Suppose the matrix $A$ is symmetric. Then we can find a basis of $n$ orthonormal eigenvectors $v_1,\cdots v_n$ with eigenvalues $\lambda_1,\dots,\lambda_n$. Each eigenvalue geometrically represents the "stretching factor" in the direction of its associated eigenvector.

The Frobenius norm of $A$ in this case is $$\|A\|_F =\sqrt{\text{Tr}(AA^t)}=\sqrt{\text{Tr}(A^2)}=\sqrt{\lambda_1^2+\cdots+\lambda_n^2}.$$ Working in the basis $v_1,\cdots v_n$, consider the box which is the image of the unit cube under $A$. The Frobenius norm is the diagonal of that box, and the determinant is the volume.

The usual norm defined as $\sup_{\|x\|=1}\|Ax\|$ corresponds to the longest side of the box.