The asymptotic behavior ($n\gg 1$) is given by (the $x$ contribution disappears and the integral becomes solvable) : $\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\Gamma\left(\frac 43\right)$
To get the next terms of the expansion, and give you some confidence in this result, let's expand the $e^{x/(3n)}$ factor in series and use (since this integral may be rewritten as a gamma integral) : $\int_0^\infty \left(\frac x{3n}\right)^k e^{\frac{-x^3}{3n}}dx=\frac {(3n)^{(1-2k)/3}}{k+1}\Gamma\left(\frac{k+4}3\right)$
Observe that we will have to divide by $(3n)^{2/3}$ at each step $k\to k+1$ (as you may see in GEdgar's answer) getting with the expansion $e^r=1+r+\frac {r^2}2+\cdots$
$\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\left(\Gamma\left(\frac 43\right)+\frac {\Gamma\left(\frac 53\right)}{2(3n)^{2/3}}+\frac 1{2\cdot 3 (3n)^{4/3}}+\rm{O}\left(n^{-2}\right)\right)$