$$\eqalign{\log(n!) &\approx \left( \ln \left( n \right) -1 \right) n+\ln \left( \sqrt {2 \pi } \right) +\dfrac{\ln \left( n \right)}{2} +\dfrac{1}{12} n^{-1}-{\dfrac {1}{
360}}\,{n}^{-3}+O \left( {n}^{-5} \right)\cr
n\ln \left( n \right) &-n+\frac{\ln \left( n \left( 1+4\,n \left( 1+2
\,n \right) \right) \right)}{6} +\frac{\ln \left( \pi \right)}{2}\cr &= \left(
\ln \left( n \right) -1 \right) n+\ln \left( \sqrt {2 \pi } \right) +\frac{
\ln \left( n \right)}{2} +\frac{1}{12}{n}^{-1}-{
\frac {1}{288}}\,{n}^{-3}+{\frac {1}{768}}\,{n}^{-4}+O \left( {n}^{-5}
\right) \cr}
$$
So the error in Ramanujan's approximation is asymptotic to $\left(\dfrac{1}{288} - \dfrac{1}{360}\right) n^{-3} = \dfrac{1}{1440} n^{-3}$.
EDIT: an even better approximation, then, would be
$n\ln \left( n \right) -n+\dfrac{\ln \left(1/30 + n \left( 1+4\,n \left( 1+2
\,n \right) \right) \right)}{6} +\dfrac{\ln \left( \pi \right)}{2}$
where the error is asymptotic to $-\dfrac{11}{11520} n^{-4}$. Thus at $n=10$
we have $\ln 10! \approx 15.1044125730755$, Ramanujan's approximation $\approx
15.1044119983597$ and the improved approximation $\approx 15.1044126589476$.