We can actually obtain exact values for $T(n)$ for all values of $n$ and not just powers of two.
Start by working with the alternate recurrence relation
$$S(n) = 7 S(\lfloor n/2 \rfloor) + 18n^2$$
where $S(0) = 0.$
Let the binary representation of $n$ be given by
$$ n = \sum_{k=0}^{\lfloor \log_2 n\rfloor} d_k 2^k.$$
Then it is not difficult to see that the exact value of $S(n)$ is given by
$$ S(n) = 18 \sum_{j=0}^{\lfloor \log_2 n\rfloor} 7^j
\left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} d_k 2^{k-j}\right)^2.$$
Now to get an upper bound on $S(n)$ consider the case where $n$ consists of all one digits, giving
$$ S(n) \le 18 \sum_{j=0}^{\lfloor \log_2 n\rfloor} 7^j
\left( \sum_{k=j}^{\lfloor \log_2 n\rfloor} 2^{k-j} \right)^2 =
\frac{441}{5} 7^{\lfloor \log_2 n\rfloor}-96 \times 4^{\lfloor \log_2 n\rfloor} +
\frac{144}{5} 2^{\lfloor \log_2 n\rfloor} -3.$$
For a lower bound, take $n$ to be a one digit followed by zeros, giving
$$ S(n) \ge 18 \sum_{j=0}^{\lfloor \log_2 n\rfloor} 7^j
(2^{\lfloor \log_2 n\rfloor-j})^2 =
42 \times 7^{\lfloor \log_2 n\rfloor} - 24 \times 4^{\lfloor \log_2 n\rfloor}.$$
We still need to account for the fact that $T(0)=1, T(1)=1$ and $T(2)=1.$ A simple calculation shows that
$$ T(n) = S(n) - \frac{197}{7} 7^{\lfloor \log_2 n\rfloor} d_{\lfloor \log_2 n\rfloor}
+ 78 \times 7^{\lfloor \log_2 n\rfloor-1} d_{\lfloor \log_2 n\rfloor-1}.$$
This formula is exact and holds for all $n\ge 3$.
Finally to get the asymptotics look at the two leading terms from the lower and upper bounds. The first is
$$\Theta\left(7^{\lfloor \log_2 n\rfloor}\right) =
\Theta(2^{\log_2 7 \log_2 n}) = \Theta(n^{\log_2 7}).$$
$T(n)$ fluctuates around this value with the coefficient at most for strings of ones
$$\frac{1}{7} \left(\frac{441}{5} - \frac{197}{7} + \frac{78}{7} \right)= \frac{356}{35}$$
and at least
$$ 42 - \frac{197}{7} = \frac{97}{7}.$$
The next term in the asymptotic expansion is
$$\Theta\left(4^{\lfloor \log_2 n\rfloor}\right) = \Theta(n^2),$$
with the coefficient between $24$ and $96.$