So using the law of logarithms, as you have already determined:
$n\sum_{i=0}^{\log_{3}(n-1)}{\log_{3}{\frac{n}{3^{i}}}}=n\left(\sum_{i=0}^{\log_{3}{(n-1)}}{\log_{3}(n)-\sum_{i=0}^{\log_{3}(n-1)}}i \log_{3}{3}\right)$
However, we have $\log_{3}{3}=1$, so the second term can simply be evaluated: $\sum_{i=0}^{\log_{3}(n-1)}i\log_{3}{3}=\sum_{i=0}^{\log_{3}(n-1)}i=\frac{\log_{3}(n-1)\times(\log_{3}(n-1)+1)}{2}$
The first term can be evaluated by noticing that $\log_{3}(n)$ is a constant, independent of $i$, and therefore can be rewritten:
$\sum_{i=0}^{\log_{3}(n-1)}\log_{3}(n)=\log_{3}(n)\sum_{i=0}^{\log(n-1)}1=\log_{3}(n)\times(\log_{3}(n-1)+1)$
Therefore the entire series can be rewritten:
$n\sum_{i=0}^{\log_{3}(n-1)}\log_{3}{\frac{n}{3^{i}}}=n\left(\log_{3}(n)\log_{3}(n-1)+\log_{3}(n)-\frac{\log_{3}(n-1)(\log_{3}(n-1)+1)}{2}\right)$