I believe the following is a counterexample. Let $\mathcal{A}$ and $\mathcal{B}$ be closed symmetric monoidal abelian categories such that the unit object $1\in\mathcal{B}$ is projective and let $F:\mathcal{A}\to\mathcal{B}$ be a non-exact strong symmetric monoidal functor which has a right adjoint $G:\mathcal{B}\to\mathcal{A}$. For instance, if $A$ is a commutative ring and $B$ is a commutative $A$-algebra which is not flat over $A$, you could have $\mathcal{A}=\mathrm{Mod}_A$ and $\mathcal{B}=\mathrm{Mod}_B$ and $F(M)=M\otimes_A B$. Let $\mathcal{C}=\mathcal{A}\times\mathcal{B}$, and equip it with the symmetric monoidal structure given by $$(M,V)\otimes (N,W)=(M\otimes N,F(M)\otimes W\oplus V\otimes F(N)\oplus V\otimes W).$$ This structure is closed, with internal hom given by $$\operatorname{hom}((M,V),(N,W))=(\operatorname{hom}(M,N)\oplus G(\operatorname{hom}(V,W)),\operatorname{hom}(F(M),W)\oplus\operatorname{hom}(V,W)).$$
In this category, the object $(0,1)$ is projective by hypothesis, but it is not flat because $(M,0)\otimes (0,1)=(0,F(M))$ and $F$ is not exact.
In the example mentioned above where $\mathcal{A}=\mathrm{Mod}_A$ and $\mathcal{B}=\mathrm{Mod}_B$ and $B$ happens to be a quotient of $A$, this construction has the following intuitive explanation. The monoidal product is defined as if $(M,V)$ were secretly the $A$-module $M\oplus V$ and the tensor product is just the ordinary tensor product of $A$-modules. However, the category itself doesn't believe that $(M,V)$ is just a single $A$-module $M\oplus V$, 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.