Do we have for all $M \in SL_n(\Bbb K)$, $\lVert M \rVert \geq 1$ when $\lVert \cdot \rVert$ is a matrix norm?

The norm of a matrix is larger than its eigenvalues. The determinant is the product of its eigenvalues. So, if the determinant of $M$ is $1$, there must be at least one eigenvalue $\lambda$ such that $|\lambda| \ge 1$, which implies $$\|M\| \ge |\lambda| \ge 1,$$ regardless of the matrix norm used.

If $\| \cdot \|$ is a matrix norm then $\rho(A) \leq \| A\|$ and $\rho(A) = 1$

If $\lVert M\rVert<1$, then, if $B$ is the closed unit ball, the volume of $M(B)$ will be smaller than the valume of $B$. But that cannot happen because, sense, $\det M=1$, the map $v\mapsto M.v$ must preserve volumes.