You’ve made a couple of mistakes: you’ve implicitly assumed that $n$ is even, and you’ve miscounted the steps needed to reduce $T(n)$.
If $n=2m$, your calculation, with the errors fixed, shows that $T(n)=T(0)+\sum_{k=1}^m\log 2k\;;$ if $n=2m+1$, it shows that $T(n)=T(1)+\sum_{k=1}^m\log(2k+1)\;.$
In each case the summation has $\lfloor n/2\rfloor$ terms, each of which is at most $\log n$, so $T(n)\le\max\{T(0),T(1)\}+\frac{n}2\log n\;;$ this clearly implies that $T(n)$ is $O(n\log n)$.
If you want to look closer, you can work out that $\sum_{k=1}^n\log 2k=\log\prod_{k=1}^m 2k=\log 2^mm!\;,$ and
$\begin{align*} \sum_{k=1}^m\log(2k+1)&=\log\prod_{k=1}^m(2k+1)\\ &=\log\Big(3\cdot5\cdot \ldots\cdot(2m+1)\Big)\\ &=\log\frac{(2m+1)!}{2\cdot4\cdot\ldots\cdot(2m)}\\ &=\log\frac{(2m+1)!}{2^mm!}\;, \end{align*}$
but these aren’t easily going to give you nice bounds better than the one that you already have. A better idea, if you need more precise information, is to note that the sums of logs can be approximated by integrals over appropriate limits of the function $f(x)=\log 2x$.