My question wasn't really precise. I knew how to compute the probability, but I didn't know how to combine it with Doob's inequality. But that's fine now, too. One just has to define $\tau_M := \inf \{t \ge \tau^-:\dots \}$ where $\tau^-$ is the first hitting time of $-\epsilon \sqrt{vl}$. Then one can can combine these results to get a lower bound on the probability of $X_l$ smaller than $-\epsilon/2 \sqrt{vl}$ conditioned on $\tau \le l$ and $X_{\tau} \le -\epsilon \sqrt{vl}$ and get the wanted result.

Thanks for your helpful answer, Serguei. I get the idea of the proof and I'm able to comprehend each of the steps formally, but I do not understand how you make sure that the probability of staying left of $-1/2\epsilon\sqrt(vl)$ conditioned on having hit $- \epsilon\sqrt(vl)$ before time $l$ is strictly positive (so how you combine steps (3)-(5)). It seems to me that you want to use some Markov-like property in step 4 (without assuming a Markov process?).