High probability events without low probability coordinates
How does $\epsilon$ compare to $n$? If $\epsilon$ can be $O(1/\sqrt{n})$, then I think we can accomplish what you want. Let $B = \mbox{Supp}(X) - E$. Note that $B$ is given $\epsilon$ probability mass under $X$. Let $\lambda(i,\sigma)\epsilon$ denote the probability mass assigned to strings in $B$ such that the $i$th coordinate has symbol $\sigma$. 

Suppose $(i,\sigma)$ were a low probability coordinate for some strings in $E$. Let $\delta(i,\sigma)$ denote the probability mass assigned to those strings. Then, by definition, $\frac{\delta(i,\sigma)}{\delta(i,\sigma) + \lambda(i,\sigma)\epsilon} \leq \epsilon$, implying that $\delta(i,\sigma) \leq 2\lambda(i,\sigma)\epsilon^2$. We can discard these low probability strings while only suffering a $\delta(i,\sigma)$ loss in prob. mass to $E$. 

Continue doing this for all possible bad $(i,\sigma)$, and in the end we only discard at most $\sum_{i,\sigma} \delta(i,\sigma) \leq \sum_{i} \sum_{\sigma} 2\lambda(i,\sigma)\epsilon^2 \leq 2\sum_{i} \epsilon^2 = 2n\epsilon^2$. This uses the fact that for all $i$, $\sum_{\sigma} \lambda(i,\sigma) = 1$. 

If you wanted $E'$ to have probability mass $1 -\gamma$, then $\epsilon$ needs to be such that $\epsilon + 2n\epsilon^2 \leq \gamma$, or that $\epsilon = O(\gamma/\sqrt{2n})$ suffices. 

It's not clear to me at the moment whether this dependence on $n$ can be gotten rid of; I'll continue thinking about it.