Example for a norm on Hom(V,W) which is not determined by rank-one operators Assume $(V,\|\cdot \|_V),(W\|\cdot \|_W)$ are two finite dimensional normed spaces (over $\mathbb{R}$). 
Any operator norm on $\text{Hom}(V,W)$ is determined by its value on rank-one operators. (This is a corollary from a reconstruction argument given here, see the end of the answer).
I suspect there are (many) norms on $\text{Hom}(V,W)$ which do not have this property. (They are not determined by evaluating only on rank-ones).
I am looking for an example. 

An operator norm on $\operatorname{Hom}(V,W)$ is determined by its value on rank-one operators once we know it's an operator norm.  
Without this additional information, it is not so determined. First, I claim that  the operators of rank $\le 1$ form a manifold  $R_1(V,W)$ of dimension $$\dim R_1(V,W) = \dim V+\dim W-1$$ 
Indeed, fix a norm on $V$ and $W$. Each rank-one functional  is represented as $v\mapsto \lambda f(v)w$ for some $\lambda\in (0,\infty)$, some unit linear functional $f\in V^*$ and some unit  vector $w\in W$. So, $R_1(V,W)$ is diffeomorphic to the product of $\mathbb{R}$  and two spheres.  This gives the dimension.
The dimension of $R_1$ is strictly less than $$\dim\operatorname{Hom}(V,W) = (\dim V)\times (\dim W)$$ unless $\dim V=\dim W = 1$. Therefore, the closure of $R_1$ has empty interior (in fact, it's just $R_1$ together with the zero operator).
The knowledge of a norm is equivalent to the knowledge of its unit sphere, which is just the boundary of a convex bounded centrally symmetric set with nonempty interior. If we only know the intersection of such a set  with $R_1(V,W)$, there are infinitely many possibilities for what it can be, e.g., near identity. 
For a concrete example: all Schatten norms $S_p$ agree on rank-one operators, because a rank-one operator has just one nonzero singular value. In particular, $S_p$ agrees with the operator norm (for the Euclidean vector norm), which is  equal to the largest singular value.
