Every norm sphere of M_n(R) is a sphere

Setup. $M_n(\mathbb{R})\cong\mathbb{R}^{n^2}$, and $\|\cdot\|_F$ is the Euclidean norm of this space, so $S_F=S^{n^2-1}$. All norms on a finite dimensional space are equivalent, hence $N$ is continuous for the usual topology, and $N(A)>0$ for $A\ne0$.

The two maps. Define
$$\varphi:S_N\to S_F,\ \ \varphi(A)=\frac{A}{\|A\|_F},\qquad\psi:S_F\to S_N,\ \ \psi(B)=\frac{B}{N(B)}.$$
They are well defined: $A\in S_N$ is nonzero so $\|A\|_F>0$, and $B\in S_F$ is nonzero so $N(B)>0$. They are continuous, being quotients of continuous functions with nonvanishing denominator.

They are inverse to each other. Take $A\in S_N$ and set $B=\varphi(A)=A/\|A\|_F$. Then $N(B)=N(A)/\|A\|_F=1/\|A\|_F$, so
$$\psi(B)=\frac{B}{N(B)}=\frac{A/\|A\|_F}{1/\|A\|_F}=A.$$
Symmetrically $\varphi\circ\psi=\mathrm{id}$.

Conclusion. $\varphi$ is a homeomorphism $S_N\to S^{n^2-1}$.

Remark. The same argument works for every norm on any $\mathbb{R}^m$: all unit spheres are homeomorphic to $S^{m-1}$, and the unit balls are homeomorphic to the closed ball. It also shows that "the" topology of $M_n(\mathbb{R})$ does not depend on the chosen norm.