From the standard mathematics of complex exponentials:
$$\begin{align} e^{-iE\,t(1-i\epsilon)} &= e^{-iE\,t}e^{-E\,\epsilon\, t} \\ &= e^{-E\,\epsilon\,t}\left(\cos(Et) + i \sin(Et)\right) \end{align}$$$$\begin{align} e^{-iE\,t(1-i\epsilon)} &= e^{-iE\,t}e^{-E\,\epsilon\, t} \\ &= e^{-E\,\epsilon\,t}\left(\cos(Et) - i \sin(Et)\right) \end{align}$$
Since, definitionally, $n=0$ is the lowest possible value for $E$, and it appears in a negative exponential in front of a term of magnitude 1, the $n=0$ state falls off most slowly for all $\epsilon > 0$