Dividing both sides by $n$ then introducing $S(m):=\frac{T(2^n)}{2^n}$ yields:
$$S(m)=S(m/2)+1+\frac{1}{m}$$
It follows that: $$1<S(m)-S(m/2)\leq 2$$ And further: $$\forall k, 1<S(m/2^k)-S(m/2^{k+1})\leq 2$$ Now summing for $k=1\dots \log(m)-1$ gives us: $$S(m)=\Theta(\log(m))$$