Allen, thanks for your input. I am a little puzzled now. In principle, I can take the Clebsch-Gordan coefficients of a tensor decomposition and use them to compute the $6j$ symbols. This amounts to choose a basis of $A\otimes B$. Tensoring the result with another representation in $C$, and computing the Clebsch-Gordan coefficients, will give me a basis of $(A\otimes B)\otimes C$. Doing the same in the opposite order gives a basis in $A\otimes(B\otimes C)$. Comparing the two should give the $6j$. A related thread in which you also participated is mathoverflow.net/q/15800/103992