Assume $(V,\| \|_V),(W,\| \|_W)$ are both finite dimensional normed spaces. We have the induced operator norm on $Hom(V,W)$.
When does it occur that this norm is actually induced from some inner product?
As observed by a comment of user225318, If $dimV=dimW=1$ then the answer can clearly be positive. It can be seen that in the case where $dimV=1 (V=\mathbb{R})$ and the norm on $W$ is induced by an inner product, the answer is positive. So let us require $dimV>1$ or that $\| \|_W$ is not induced by an inner product.
To state this more clearly, I would be happy to find a complete characterization, i.e necessary & sufficient conditions on the dimensions of $V,W$ and on their respective norms that are equivalent to the operator norm being induced by an inner product.
I assume here the spaces are finite dimensional (as in the question), and the vector spaces are real (for convenience, one can generalize to other fields too).
Recall first that a norm is induced from an inner product if and only if it satisfies the polarization identity: $$ 2\|u\|^2 + 2\|v\|^2 = \|u+v\|^2 + \|u-v\|^2 $$ for every $u,v\in V$.
Let $V, W$ be normed vector spaces.
Claim. If the norm on $W$ is not induced by an inner product, then the operator norm on $L(V,W)$ is not induced by an inner product.
Proof. Let $\phi:V \to \mathbb{R}$ be a linear functional such that $\sup_{v\in V\setminus \{0\}} |\phi(v)| / \|v\| = 1$. Take $w_1\neq w_2\in W$. The operators $w_j \phi$ are both in $L(V,W)$. It is easy to see that the operator norm $\|w_j \phi\|_{L(V,W)} = \|w_j\|_W$, and so the operator norm is induced by an inner product only if the norm on $W$ is induced by an inner product. Q.E.D.
In other words, one can isometrically embed $W \simeq L(\mathbb{R},W)$ into $L(V,W)$. So if the latter is an inner product space so is the former.
Claim. If the norm on $V$ is not induced by an inner product, then the operator norm on $L(V,W)$ is not induced by an inner product.
Proof. Fix $w_0\in W$. Then $L(V,\mathbb{R})$ embeds in $L(V,W)$ via $\phi \mapsto w \phi$. Hence if the norm on $L(V,W)$ is induced by an inner product, so must the operator norm on $L(V,\mathbb{R})$ which is the same as the dual space of $V$. This implies that $V'$ is an inner product space and hence so is $V$. Q.E.D.
In other words, one can isometrically embed $V' \simeq L(V,\mathbb{R})$ into $L(V,W)$, so if the latter is inner product, so is the former.
Claim. If $\dim V, \dim W > 1$ and $V,W$ are inner product spaces, the operator norm $L(V,W)$ is not induced by an inner product.
Proof. Let $v_1 \perp v_2 \in V$ and $w_1\perp w_2 \in W$ be unit vectors. Consider the mappings $u_1 = v \mapsto \langle v, v_1\rangle w_1$ and $u_2 = v \mapsto \langle v, v_2\rangle w_2$. One easily calculate that the operator norm $\|u_1\| = \|u_2\| = 1$. One calculates further that the operator norm $\|u_1 + u_2\| = \|u_1 - u_2\| = 1$. This violates the polarization identity. Q.E.D.
To conclude, $L(V,W)$ is an inner product space with respect to the operator norm iff both $V, W$ are inner product spaces and at least one of $V, W$ has dimension $1$.