The trace of the standard representation $\Psi$ is just the number $Fix(\pi)$ of fixed points of the permutation $\pi$. Now if $A\in\Hom(E\oplus F)$ and $A(E)\sbe E$ and $A(F)\sbe F$, then $\tr A=\tr A|E+\tr A|F$. In our case $F=[e_1+\cdots+e_n]$ and $E=F^\perp$, so $$ \tr\Psi(\pi)|E =\tr\Psi(\pi)-\tr\Psi(\pi)|F =Fix(\pi)-1~. $$