$\DeclareMathOperator{\Jac}{\mathrm{Jac}}\DeclareMathOperator{\Sym}{\mathrm{Sym}}$
The answer is yes.

First of all let's see how to identify the tangent space of a point $[\Jac C] \in A_g$ with $\Sym^2 H^1(C,O_C)$.

The 1st order deformations of an abelian variety $A$ (not necessarily preserving the polarization) are given by $H^1(A,T_A) \cong H^1(A,O_A) \otimes H^0(A,\Omega_A)^\vee$ since $T_A$ is the trivial bundle $O_A \otimes H^0(A,\Omega_A)^\vee$. (You get a trivialization of $T_A$ by translation on the group $A$, identifying each fiber canonically with the tangent space at the origin.) The map $H^1(C,\mathbb C) \to H^1(\Jac C, \mathbb C)$ is an isomorphism of Hodge structures, so for a Jacobian the latter deformation space can be rewritten as $H^1(C,O_C) \otimes H^0(C,\Omega_C)^\vee$. Serre duality (on $C$!) identifies $H^0(C,\Omega_C)^\vee$ with $H^0(C,O_C)$, so for a Jacobian the deformations can be written as $H^1(C,O_C) \otimes H^1(C,O_C)$. Finally $\Sym^2 H^1(C,O_C)$ sits inside the deformation space as the deformations preserving the polarization on $\Jac C$ (and hence the isomorphism $H^1(A,O_A)\cong H^1(A,\Omega_A)^\vee$).

Now cup product defines a map $H^1(C,T_C) \otimes H^0(C,\Omega_C) \to H^1(C,O_C)$, which by dualizing gives $H^1(C,T_C) \to H^1(C,O_C) \otimes H^0(C,\Omega_C)^\vee$. The claim is that this is the differential of the Torelli map. A proof can be found in Arbarello, Cornalba, Griffiths "The geometry of algebraic curves vol. 2" in the chapter on deformation theory, section 8.

Transposing and applying Serre duality gives also a map $H^0(C,\Omega_C) \otimes H^0(C,\Omega_C) \to H^0(C, \Omega_C^{\otimes 2})$ -- the codifferential of Torelli -- and what we want to show is that this map, too, is given by cup product. So what you need to know is that Serre duality is compatible with cup product maps. But this is immediate if you think of Serre duality as *defined* by cup products taking values in $H^1(C,\Omega_C)$ with its canonical isomorphism with $\mathbb C$ defined via the trace map.