We can solve another closely related recurrence that admits an exact
solution and makes it possible to get precise bounds. Suppose we have
$T(0)=0$ and $T(1)=1$ and for $n\ge 2$
$$T(n) = T(\lfloor n/2\rfloor) +
q\times n\times\lfloor \log_2 n\rfloor.$$
Furthermore let the base two representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Then we can unroll the recurrence to obtain the following exact
formula for $n\ge 2$
$$T(n) = 1 + q\sum_{j=0}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor - j)
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
This follows more or less by inspection.
Now to get an upper bound consider a string of one digits, which gives
$$T(n) \le
1 + q\sum_{j=0}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor - j)
\sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}
\\= 1 +
q\left(
4 \lfloor \log_2 n \rfloor 2^{\lfloor \log_2 n \rfloor}
- 4 \times 2^{\lfloor \log_2 n \rfloor}
- \frac{1}{2} (\lfloor \log_2 n \rfloor)^2
- \frac{1}{2} \lfloor \log_2 n \rfloor
+ 4\right).$$
For a lower bound consider a one followed by a string of zero digits,
giving
$$T(n) \ge
1 + q\sum_{j=0}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor - j)
2^{\lfloor \log_2 n \rfloor-j}
\\ =
1 + q\left(
2 \lfloor \log_2 n \rfloor 2^{\lfloor \log_2 n \rfloor}
- 2 \times 2^{\lfloor \log_2 n \rfloor}
+ 2
\right).$$
Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$\color{#006}{q \lfloor \log_2 n \rfloor 2^{\lfloor \log_2 n \rfloor}
\in \Theta\left(\log_2 n \times 2^{\log_2 n}\right)
= \Theta\left(\log n \times n\right)}.$$
These are in agreement with what the Master theorem would produce.
Here is another computation in the same spirit:
MSE link.
