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 for $n\ge 1$ (this gives $T(1)=1$)
$$T(n) = n + \sum_{q=1}^p T(\lfloor n/2^q \rfloor).$$
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 $n\ge 1$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
This follows more or less by inspection.
Now to get an upper bound consider a string of one digits which yields
$$T(n) \le \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}
\\= \sum_{j=0}^{\lfloor \log_2 n \rfloor}
(2^{\lfloor \log_2 n \rfloor+1-j}-1)
[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\\= 2^{\lfloor \log_2 n \rfloor+1}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(\frac{1}{2} \right)^j[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
- \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\sum_{q=1}^p z^q}.$$
Note that this bound is attained and cannot be improved.
Since $(1/2)^j \le 1$ it is upper bounded by
$$(2^{\lfloor \log_2 n \rfloor+1} - 1)
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(\frac{1}{2} \right)^j[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\\ \le (2^{\lfloor \log_2 n \rfloor+1} - 1)
\sum_{j=0}^\infty
\left(\frac{1}{2} \right)^j[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\\= (2^{\lfloor \log_2 n \rfloor+1} - 1)
\frac{1}{1-\sum_{q=1}^p (1/2)^q}
\\= 2^p (2^{\lfloor \log_2 n \rfloor+1} - 1).$$
The lower bound is for the case of a one digit followed by a string of
zeros and yields
$$T(n) \ge \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
2^{\lfloor \log_2 n \rfloor-j}
\\= 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(\frac{1}{2} \right)^j[z^j] \frac{1}{1-\sum_{q=1}^p z^q}.$$
At this point unfortunately we cannot avoid studying the roots of
$$f(z) = 1-\sum_{q=1}^p z^q.$$ We need the root $\rho$ that is closest
to the origin and more precisely, we have to show that $\rho > 1/2.$
Since $f(0) = 1$ and $f(1) = - (p-1)$ we have at least one root $\rho$
in $(0, 1)$ and by the sign and the continuity of $f'(z)$ it is the
only one. We must have $\rho > 1/2$ since otherwise $\sum_{q=1}^p z^q
< 1$ because $p$ is finite. Furthermore there are no complex roots
with modulus $\rho$ because by the triangle inequality there would be
cancellation and we would again have $|\sum_{q=1}^p z^q| < 1.$ This last
argument also applies to the negative real negative root that appears
when $p$ is even.
The conclusion is that $\rho$ is the dominant singularity and
$$[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\sim -\frac{1}{\rho} \mathrm{Res}(1/f(z); z=\rho) \times \rho^{-j}
\\= \frac{1}{\sum_{q=1}^p q\times \rho^{q-1}} \times \rho^{-(j+1)}
= \frac{1}{\sum_{q=1}^p q\times \rho^q} \times \rho^{-j}.$$
Here we have used the fact that
$$\frac{1}{z-\rho} = - \frac{1}{\rho} \frac{1}{1-z/\rho}.$$
We have shown that
$$\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(\frac{1}{2} \right)^j[z^j] \frac{1}{1-\sum_{q=1}^p z^q}
\sim \frac{1}{\sum_{q=1}^p q\times \rho^q}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(\frac{1}{2\rho} \right)^j.$$
Now use the fact that $\rho > 1/2$ to get $1 > 1/2/\rho$ to obtain
(geometric series)
that this sum is bounded above by a constant that does not depend on
$n.$
We are now ready to conclude. We have established that
- $T(n)$ is upper bounded by
$2^p (2^{\lfloor \log_2 n \rfloor+1} - 1)$
- $T(n)$ is lower bounded by a multiple of
$2^{\lfloor \log_2 n \rfloor}$ times a coefficient from
an interval that does not depend on $n.$
Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$\color{#006}{2^{\lfloor \log_2 n \rfloor}
\in \Theta\left(2^{\log_2 n}\right)
= \Theta\left(n\right)}.$$
These are both in agreement with what the Master theorem would produce.
Remarks. This MSE link I and this MSE link II present closely related arguments. The calculation of the upper bound uses an annihilated coefficient extractor which is also used at this MSE link III.
