I just accidentally stumbled upon this nice question. I suspect that by now you know the answer yourself, but still, let me do one simple computation. If you like it, I'll think more of the question.
The process in question is just the following. A particle departs from the origin and does i.i.d. random steps with positive mean. If it ever ends up to the right of the origin, it is put back there. The question is what is the leftmost position it visited after $t$ steps.
I want to argue as follows. Let's look at the probability that it'll reach $-D$ before it comes back to the origin. It is about $Ce^{-\lambda D}$ where $\lambda>0$ is the only real positive solution of the equation $\int p(x)e^{\lambda x}dx=0$ p(x)e^{-\lambda x}dx=1$ and $p$ is the density of the step distribution. The value of $C$ is also possible to compute. Now, each attempt to depart from the origin lasts for some time with exponentially decaying tails. Let $T$ be the average time of travel. Then, by the time $t$, the number of attempted departures is about $t/T$. Thus, the probability of success is about $(1-Ce^{-\lambda D})^{t/T}$ meaning that $ED_{\text{min}}\approx \lambda^{-1}(\log t+\log(C/T)+U)$ where $U$ is some universal constant ($U=\int_0^\infty (e^{-e^{-x}}+e^{-e^x}-1)dx$, if I haven't mistaken). This would mean that, for large times, you are always a constant number of times off with the Brownian approximation.
Of course, this is just a back of envelope computation, but, since I don't even know if you are still interested, I'd rather stop here.
