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This is false for other permutations. For example, call $(e_1,e_2,e_3)$ the canonical basis of $\mathbb{R}^3$ and let A,B,C be the $3 \times 3$ real matrices such that $Ae_1=e_2$, $Be_2=e_3$, $Ce_3=e_1$, and the other images of the $e_j$ by $A,B,C$ are null. Then $BC$ is null, so $ABC$ is null. Yet $CBAe_1=e_1$ and $CBAe_j=0$ for $j \in \{1,2\}$, so the trace of $CBA$ is $1$.