Sum of product of three characters over group elements

Let $$G$$ be a finite group, and $$\{\chi_i\}$$ the set of characters of its complex irreducible representations. I'm trying to compute $$\sum_{g \in G} \chi_i (g) \chi_j (g) \chi_k (g)$$ for any three irreducible characters $$\chi_i, \chi_j, \chi_k$$.

According to this question, it can be interpreted as

the multiplicity of the trivial rep of $$G$$ in the (tensor) product rep $$\chi_1\otimes \chi_2\otimes \chi_3$$ of $$G\times G\times G$$ restricted to $$G$$ via $$g \mapsto (g,g,g)$$.

From this, if the restricted representation is irreducible, then the sum should be $$|G|$$ if the restricted representation is trivial and $$0$$ otherwise.

However, this is not really simpler than computing the above sum from the character table. Is there a simple way of computing the above sum?

Indeed, this sum is $$|G|$$ times the multiplicity of the trivial representation inside the tensor product $$\chi_1 \otimes \chi_2 \otimes \chi_3$$ of the three representations. It is also the multiplicity of $$\bar{\chi}_3$$ (i.e. the dual character to $$\chi_3$$) inside $$\chi_1 \otimes \chi_2$$.
These are called Clebsch-Gordan coefficients. As far as I understand while we have this nice formula as a character sum (so you can just compute them for any particular group), they are otherwise pretty difficult to understand in general. In the case where $$G$$ is a symmetric group these give the so-called Kronecker coefficients, which are famously difficult to understand combinatorially.