Is the matrix positive definite given the Gauss-Seidel method converges?

There is an interesting though partial answer to your question:

Under your assumptions, it is not possible that the signature of $A$ be $(n-1,1)$ (exactly one negative eigenvalue).

Proof: With standard notations, $A=D-E-E^T$ where $E$ is strictly triangular and $D$ diagonal. By assumption, $D>0_n$. Notice that the assumption of convergence implies that $A$ is invertible.

The iteration matrix is $G=(D-E)^{-1}E^T$. The quadratic form $q(x):=x^TAx$ satisfies the identity $$q(Gx)+y^TDy=q(x),\qquad y:=x-Gx=(D-E)^{-1}Ax.$$ This implies that $G$ preserves the set $q<0$. If the signature of $A$ is $(n-1,1)$, this set is the union of two opposite convex cones $\pm K$. Thus $\pm G$ preserves $K$. One deduces from Brouwer fixed point theorem that $G$ admits an eigenvector $x\in K$, $Gx=\mu x$. Then $\mu^2q(x)<q(x)<0$ gives $|\mu|>1$, which contradicts the convergence of the Gauss-Seidel method.

Nota. The identity about $q$ is the one used to prove the convergence of the method when $A$ is positive definite. More generally, it is used to prove that any method with iteration matrix $L:=M^{-1}N$ where $A=M-N$, such that both $A$ and $M^T+N$ are positive definite, is convergent