Your expression actually means
$\displaystyle \displaystyle\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3 \epsilon_{ijk} \frac{\partial ^2 A_{k}}{\partial x_{i}\partial x_{j}}$
This is
$\displaystyle \epsilon_{111}\frac{\partial^2 A_{1}}{\partial x_{1}^2}+\epsilon_{112}\frac{\partial ^2 A_{1}}{\partial x_{1}\partial x_{2}}+\epsilon_{113}\frac{\partial ^2 A_{1}}{\partial x_{1}\partial x_{3}}+\epsilon_{121}\frac{\partial ^2 A_{1}}{\partial x_{2}\partial x_{1}}+\epsilon_{122}\frac{\partial ^2 A_{2}}{\partial x_{1}\partial x_{2}}+...$
The definition of $\displaystyle \epsilon_{ijk}$ gives you a lot of zeros, in particular the only non zero are $\displaystyle 1=\epsilon_{123}=\epsilon_{231}=\epsilon_{312}$ and $\displaystyle -1=\epsilon_{321}=\epsilon_{132}=\epsilon_{213}$. From this fact and that
$\displaystyle \frac{\partial ^2}{\partial x_{i} \partial x_{j}}=\frac{\partial ^2}{\partial x_{j} \partial x_{i}}$
you will have pairs of terms in your sum like
$\displaystyle \epsilon_{123}\frac{\partial A_{3}}{\partial x_{1}\partial x_{2}}+\epsilon_{213}\frac{\partial A_{3}}{\partial x_{2}\partial x_{1}}$
which cancel and gives you zero.