It is not an answer to your question, but I hope it will help:

The way of doing Fourier Transform in abelian group $A$ can be described in the is the following way: Let $f,g \in F[A]$ We know that $F[A]$ is isomorphic to the space $F^A$ with pointwise multiplication. Let $T$(which is acctually Fourier Transform) be this isomorphism. If we want calculate $f*g$ then calculate $T^{-1}(T(f)\cdot T(g))$. In case of non abelian group like $S_n$ It holds that $F[G]$ is isomorphic to the direct sum of matrix algebras that is $F[G]\simeq\oplus M_{n_i}$. Thus using the same formula you can calculate convolution in $S_n$, but now you will need to multiply matrixes.

It is not an answer to your question, but I hope it will help:

In general arithmetic complexity of convolution in non-anelian groups "equivalent" to the complexity of matrix multiplication. Here is the reason why:

The way of doing Fourier Transform in abelian group $A$ can be described in the is the following way: Let $f,g \in F[A]$ We know that $F[A]$ is isomorphic to the space $F^A$ with pointwise multiplication. Let $T$(which is acctually Fourier Transform) be this isomorphism. If we want calculate $f*g$ then calculate $T^{-1}(T(f)\cdot T(g))$. In case of non abelian group like $S_n$ It holds that $F[G]$ is isomorphic to the direct sum of matrix algebras that is $F[G]\simeq\oplus M_{n_i}$. Thus using the same formula you can calculate convolution in $S_n$, but now you will need to multiply matrixes.