$A \wedge A \wedge A$ in Chern-Simons

Option (1) Use the definition $(\omega \otimes S) \wedge (\eta \otimes S) = (\omega \wedge \eta) \otimes (S\otimes T)$ of the wedge product for Lie algebra valued forms. Define Lie bracket and Killing form as bilinear maps $[S\otimes T] = [S,T]$ and $\langle S \otimes T \rangle = \langle S, T\rangle$. Then the formula that you want is $$\langle A \wedge [A \wedge A] \rangle,$$ where the commutator and Killing form apply only to the Lie algebra factors, ignoring the differential form factors.

Option (2) Use the definition $(\omega \otimes S) \wedge (\eta \otimes S) = (\omega \wedge \eta) \otimes ST$ of the wedge product of forms valued in a particular matrix representation of a Lie algebra. Then the formula that you want is $$\operatorname{tr} (A \wedge A \wedge A),$$ where again the trace applies only to the matrix factors ignoring the differential form factors.

The two formulas agree up to a constant factor, as long as your Lie algebra is simple.

For Lie algebras of matrices (which is what you really care about in Chern-Simons theory) think of $A$ as a form with matrix coefficients

$$ A=\sum_i A_i dx^i, $$

where $A_i$ are $r\times r$ matrices.With this convention, use the usual wedge product

$$\left(\sum_i A_i dx^i\right)\wedge \left(\sum_j A_j dx^j\right)\wedge \left(\sum_k A_k dx^k\right) = \sum_{i,j,k} A_iA_jA_k dx^i\wedge dx^j\wedge dx^k, $$

where you need to recall that the product of matrices is not commutative.