Since this question is asked frequently I will try to work out a solution for generic positive integers $a$ where $a\ge 2$.
Suppose we have $T(0)=0$ and $T(1)=T(2)=\ldots =T(a-1)=1$ and for $n\ge a$ $T(n) = a T(\lfloor n/a \rfloor) + n \lfloor \log_a n \rfloor.$
Furthermore let the base $a$ representation of $n$ be $n = \sum_{k=0}^{\lfloor \log_a n \rfloor} d_k a^k.$
Then we can unroll the recurrence to obtain the following exact formula for $n\ge a$ $T(n) = a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times \sum_{k=j}^{\lfloor \log_a n \rfloor} d_k a^{k-j}.$
Now to get an upper bound consider a string consisting of the digit $a-1$ to obtain $T(n) \le a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times \sum_{k=j}^{\lfloor \log_a n \rfloor} (a-1) \times a^{k-j}.$
This simplifies to $a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times (a-1) \sum_{k=0}^{\lfloor \log_a n \rfloor-j} a^k$ which is $a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) (a^{\lfloor \log_a n \rfloor + 1 -j} -1)$ which turns into $a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} (\lfloor \log_a n \rfloor - j) (a^{\lfloor \log_a n \rfloor + 1} - a^j).$ The sum produces four terms.
The first is $\lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor + 1}.$ The second is $- \lfloor \log_a n \rfloor \frac{a^{\lfloor \log_a n \rfloor}-1}{a-1}.$ The third is $- \frac{1}{2} a^{\lfloor \log_a n \rfloor + 1} (\lfloor \log_a n \rfloor -1) \lfloor \log_a n \rfloor$ and the fourth is $\frac{1}{(a-1)^2} \left(a + a^{\lfloor \log_a n \rfloor} (\lfloor \log_a n \rfloor (a-1) -a)\right).$
This bound represented by these four terms plus the leading term is actually attained and cannot be improved upon. For the asymptotics we only need the dominant term, which is $\left(a - \frac{1}{2} a \right) \lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor} = \frac{1}{2} a \lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor}.$
Now for the lower bound, which occurs with a one digit followed by zeroes to give $T(n) \ge a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times a^{\lfloor \log_a n \rfloor-j}.$ This simplifies to $a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} (\lfloor \log_a n \rfloor - j) \times a^{\lfloor \log_a n \rfloor}$ which is $a^{\lfloor \log_a n \rfloor} + a^{\lfloor \log_a n \rfloor} \sum_{j=0}^{\lfloor \log_a n \rfloor -1} (\lfloor \log_a n \rfloor - j)$ which finally produces $a^{\lfloor \log_a n \rfloor} + a^{\lfloor \log_a n \rfloor} \sum_{j=1}^{\lfloor \log_a n \rfloor} j$ or $a^{\lfloor \log_a n \rfloor} + \frac{1}{2} \lfloor \log_a n \rfloor (\lfloor \log_a n \rfloor +1) a^{\lfloor \log_a n \rfloor}.$ The dominant term here is $\frac{1}{2} \lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor}.$
Joining the dominant terms of the upper and the lower bound we obtain the asymptotics $\color{#00A}{\lfloor \log_a n \rfloor^2 \times a^{\lfloor \log_a n \rfloor} \in \Theta\left((\log_a n)^2 \times a^{\log_a n}\right) = \Theta\left((\log n)^2 \times n\right)}.$
This MSE link has a series of similar calculations.