Let $z$ be the vector with all entries equal to 1. If $Az=0$ and $B=A+A^T$, then
$z^TBz=0$. Choose an orthogonal basis for $\mathbb{R}^n$ with $z$ as its first vector. If $C$ is the matrix representing $B$ relative to this basis, then $C_{1,1}=0$. Also $B$ is positive semidefiniteif and only if $C$ is.

If $C$ is positive semidefinite and $C_{1,1}=0$, all entries in the first row and column of $C$ must be zero. Let $C_1$ be the matrix we get by deleting the first row and column of $C$. Then $C$ is positive semidefinite if and only if the entries in its first row and column are zero and $C_1$ is positive semidefinite.

So we can reduce the question of whether an $n\times n$ matrix of the form $A+A^T$ is positive semidefinite to deciding whether an $(n-1)\times(n-1)$ matrix is positive semidefinite.

If $B$ is positive semidefinite and $z^TBz=0$, then $Bz=0$ and also $A^Tz=0$.
So in terms involving only $B$, we can say that $B$ is positive semidefinite
if and only if $Bz=0$ and the matrix representing the action of $B$ on the orthogonal complement to the span of $z$ is positive semidefinite.

The brief summary seems to be that knowing that $B=A+A^T$ with $Az=0$ does not give you much.