You may be interested to know that the recurrence
$$ T(n) = 8\;T(\lfloor n/2 \rfloor) + 1/4 \; n^2 $$
where $T(0) = 0$ and $T(1) = 1/4$ actually has a closed form solution for all $n$ and not just for $n$ a power of two. (We will show how to solve this with $T(1) = 1$ at the end of this post.)
This closed form is obtained by unrolling the recursion according to the binary digits of $n.$ Let those digits be given by
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Then the closed form solution (exact for all $n$) is given by
$$ T(n) = \frac{1}{4} \sum_{k=0}^{\lfloor \log_2 n \rfloor} 8^k
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} d_j 2^{j-k} \right)^2.$$
To get an upper bound consider the case of all one digits, which gives
$$ T(n) \le \frac{1}{4} \sum_{k=0}^{\lfloor \log_2 n \rfloor} 8^k
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} 2^{j-k} \right)^2 =
\frac{20}{21} 8^{\lfloor \log_2 n \rfloor}
- 4^{\lfloor \log_2 n \rfloor}
+ \frac{1}{3} 2^{\lfloor \log_2 n \rfloor}
- \frac{1}{28}.$$
This bound is attained.
To get a lower bound consider the case of a one digit followed by a string of zeros, giving
$$ T(n) \ge \frac{1}{4} \sum_{k=0}^{\lfloor \log_2 n \rfloor} 8^k
\left( 2^{\lfloor \log_2 n \rfloor-k} \right)^2
= \frac{1}{2} 8^{\lfloor \log_2 n \rfloor}
- \frac{1}{4} 4^{\lfloor \log_2 n \rfloor}.$$
This bound too is attained.
Joining these two it now becomes evident that
$$T(n)\in \Theta\left(8^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(2^{3\lfloor \log_2 n \rfloor}\right)
= \Theta(n^3).$$
The above calculation has $T(1) = 1/4$ while the query asks for $T(1) = 1$. It is not difficult to see that the answer to the original query can be obtained by taking
$$ T(n) + \frac{3}{4} 8^{\lfloor \log_2 n \rfloor}.$$
This gives the following upper and lower bounds
$$ \frac{143}{84} 8^{\lfloor \log_2 n \rfloor}
- 4^{\lfloor \log_2 n \rfloor}
+ \frac{1}{3} 2^{\lfloor \log_2 n \rfloor}
- \frac{1}{28}
\quad\text{and}\quad
\frac{5}{4} 8^{\lfloor \log_2 n \rfloor}
- \frac{1}{4} 4^{\lfloor \log_2 n \rfloor}.$$
The asymptotic complexity is not affected.
There is more material on this method here.
