I believe the following is a counterexample. Let $A$ be a commutative ring and $B$ be a commutative $A$-algebra which is not flat over $A$. Let $\mathcal{C}=\mathrm{Mod}_A\times \mathrm{Mod}_B$ (also known as $\mathrm{Mod}_{A\times B}$), and equip it with the symmetric monoidal structure given by $$(M,V)\otimes (N,W)=(M\otimes_A N,M\otimes_A W\oplus V\otimes_A N\oplus V\otimes_B W).$$ This structure is closed, with internal hom given by $$\operatorname{hom}((M,V),(N,W))=(\operatorname{Hom}_A(M,N)\oplus\operatorname{Hom}_B(V,W),\operatorname{Hom}_A(M,W)\oplus\operatorname{Hom}_B(V,W)).$$ In this category, the object $(0,B)$ is clearly projective, but it is not flat because $(M,0)\otimes (0,B)=(0,M\otimes_A B)$ and $B$ is not flat over $A$. Morally, what seems to be going on here is that the monoidal structure believes that $(M,V)$ is secretly the $A$-module $M\oplus V$ (at least in the case that $B$ is a quotient of $A$), but the category itself doesn't know this (and in particular doesn't have any maps from $(A,0)$ to $(0,B)$ that would break the projectivity of $(0,B)$).
For a finite $\mathbb{C}$-linear version of this example, you can take $B=\mathbb{C}$ and $A$ to be an augmented finite-dimensional $\mathbb{C}$-algebra and restrict to finitely generated modules everywhere.