Given a finite Abelian group $G$, I am interested in the sum $$ \sum_{\chi} \chi^2(g) $$ for some element $g \in G$ and the sum runs over the characters of $G$.
I tried using the identity $\chi^2(g) = \chi(g^2)$ and changing the above into a usual character sum. Is this correct or am I missing something?
In general, the Second Orthogonality Relation tells you for a general finite group $G$ and $g,h \in G$ $$\sum_{\chi \in Irr(G)}\chi(g)\overline{\chi(h)}= 0 \text{ if }g \nsim h \text{ and } = |C_G(g)| \text{ if } g \sim h. $$ (Here $\sim$ denotes conjugation) Hence if $G$ is abelian and $\chi$ linear, we have $$\sum_{\chi \in Irr(G)} \chi^2(g)=\sum_{\chi \in Irr(G)}\chi(g)\overline{\chi(g^{-1})}=0 \text{ if } g \neq g^{-1} \text{and } =|G| \text{ if } g^2=1.$$ If you want to do some calculations: see for example here (character table of $C_6$), and try an element of order $2$ and of order $3$).
There is even a generalization to groups $G$ not necessarily being abelian. If $g \in G$ and $g \sim g^{-1}$, then $g$ is called a real element. The trivial element (identity) is obviously real, also all elements of order $2$ of the center $Z(G)$. One has $$\sum_{\chi \in Irr(G)} \chi^2(g)=0 \text{ if } g \text{ is not real } \text{ and } =|C_G(g)| \text{ if } g \text { is real }.$$ If $|G|$ is odd, then $G$ does not contain any non-trivial real elements. See also here.