We let the binary representation of $n$ be
$$\sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
and assume for the base case that $T(0)=0$ and for the recursive step
$$ T(n) = T(\lfloor n/2 \rfloor)+T(\lfloor n/4 \rfloor)+T(\lfloor n/8 \rfloor)+n.$$
Then the value of $T(n)$ for all $n$ is given by
$$\sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\frac{1}{2}z -\frac{1}{4}z^2 -\frac{1}{8}z^3}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
This formula is exact for all $n.$
Now let $\rho_{1,2,3}$ be the roots of $1-\frac{1}{2}z -\frac{1}{4}z^2 -\frac{1}{8}z^3 = 0,$
with $\rho_1$ the real root. Introduce
$$ \rho =\sqrt[3]{17+3\sqrt{33}}$$
Then we have
$$\rho_1 = \frac{2}{3} \rho - \frac{4}{3} \frac{1}{\rho} -\frac{2}{3}
\quad\text{and}\quad
\rho_{2,3} = -\frac{1}{3} \rho + \frac{2}{3} \frac{1}{\rho}
-\frac{2}{3}
\pm i \sqrt{3}\left(\frac{1}{3} \rho + \frac{2}{3} \frac{1}{\rho}\right).$$
This yields
$$[z^j] \frac{1}{1-\frac{1}{2}z -\frac{1}{4}z^2 -\frac{1}{8}z^3}\\ =
\frac{8\rho_1^{-j-1}}{(\rho_1-\rho_2)(\rho_1-\rho_3)}+
\frac{8\rho_2^{-j-1}}{(\rho_2-\rho_1)(\rho_2-\rho_3)}+
\frac{8\rho_3^{-j-1}}{(\rho_3-\rho_1)(\rho_3-\rho_2)}.
$$
The numeric values of the inverses of the roots are
$$\frac{1}{\rho_1} \sim 0.9196433771
\quad\text{and}\quad
\frac{1}{\rho_{2,3}} \sim -0.2098216888 \mp 0.3031453647 i.$$
Note that $1/\rho_1$ is real and that $1/\rho_{2,3}$ has the following modulus:
$$ \left|\frac{1}{\rho_{2,3}}\right| = 0.3686763530.$$
This implies that the contribution from $1/\rho_1$ dominates asymptotically.
To see this, simply note that
$$\frac{|(1/\rho_{2,3})^j|}{|(1/\rho_1)^j|} =
\frac{|1/\rho_{2,3}|^j}{|1/\rho_1|^j}
\rightarrow 0 \quad \text{as}\quad j \rightarrow \infty.$$
What we have done here is really nothing more than apply the known
rule that the radius of convergence of a Taylor series centered at
zero is the distance to the nearest singularity, with the dominant
asymptotics of the coefficients being contributed by powers of the
inverse of the modulus of that singularity.
We are now ready to check the bounds, restricting ourselves to the
contribution from $1/\rho_1.$ The case of a lower bound occurs when
$n$ consists of a single one, followed by zeros, giving
$$ T(n) \ge 2^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\frac{1}{2}z -\frac{1}{4}z^2 -\frac{1}{8}z^3} \\ \sim
2^{\lfloor \log_2 n \rfloor}
\frac{8}{\rho_1 (\rho_1-\rho_2)(\rho_1-\rho_3)}
\sum_{j=0}^{\lfloor \log_2 n \rfloor} \rho_1^{-j}$$
or
$$\frac{8\times 2^{\lfloor \log_2 n \rfloor} }{\rho_1 (\rho_1-\rho_2)(\rho_1-\rho_3)}
\frac{1-1/\rho_1^{\lfloor \log_2 n \rfloor+1}}{1-1/\rho_1}\\=
\frac{8\times 2^{\lfloor \log_2 n \rfloor} (1-1/\rho_1^{\lfloor \log_2 n \rfloor+1}) }
{(\rho_1-1) (\rho_1-\rho_2)(\rho_1-\rho_3)}.$$
Now let
$$
P(n) = \frac{8\times 2^{\lfloor \log_2 n \rfloor}}
{(\rho_1-\rho_2)(\rho_1-\rho_3)}
\quad\text{and}\quad
Q(n) = \frac{8\times (2/\rho_1)^{\lfloor \log_2 n \rfloor}}
{(\rho_1-\rho_2)(\rho_1-\rho_3)},$$
so that the lower bound becomes
$$\frac{P(n)}{\rho_1-1} - \frac{Q(n)}{\rho_1(\rho_1-1)}.$$
The case of the upper bound occurs when $n$ is string of one digits, giving
$$T(n) \le \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\frac{1}{2}z -\frac{1}{4}z^2 -\frac{1}{8}z^3}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^k \\=
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\frac{1}{2}z -\frac{1}{4}z^2 -\frac{1}{8}z^3}
(2^{\lfloor \log_2 n \rfloor+1} - 2^j)$$
The first part of the difference is twice the term we calculated for
the lower bound. The second part is
$$- \frac{8}{\rho_1 (\rho_1-\rho_2)(\rho_1-\rho_3)}
\sum_{j=0}^{\lfloor \log_2 n \rfloor} (2/\rho_1)^j =
- \frac{8}{\rho_1 (\rho_1-\rho_2)(\rho_1-\rho_3)}
\frac{(2/\rho_1)^{\lfloor \log_2 n \rfloor+1}-1}{2/\rho_1-1}$$
or
$$- \frac{8\times (2/\rho_1)^{\lfloor \log_2 n \rfloor+1}-1}
{(2-\rho_1)(\rho_1-\rho_2)(\rho_1-\rho_3)} =
- \frac{2 Q(n)}{\rho_1(2-\rho_1)} +
\frac{1}{(2-\rho_1)(\rho_1-\rho_2)(\rho_1-\rho_3)}
$$
It follows that the upper bound is
$$ 2\frac{P(n)}{\rho_1-1}
- 2 Q(n) \left(\frac{1}{\rho_1(\rho_1-1)} + \frac{1}{\rho_1(2-\rho_1)}\right)
+ \frac{1}{(2-\rho_1)(\rho_1-\rho_2)(\rho_1-\rho_3)}.$$
Putting the upper and the lower bound together and observing that we have
$$P(n) \in \Theta(n)\quad\text{and}\quad
Q(n) \in \Theta\left((2/\rho_1)^{\lfloor \log_2 n \rfloor}\right)
= \Theta(n/2^{\log_2\rho_1 \log_2 n})
= \Theta(n^{1-\log_2\rho_1})$$
so that $$T(n) \in \Theta(n)$$ and the next term in the asymptotic expansion is
$\Theta(n^{1-\log_2\rho_1})$.
Taking certain liberties with the precise meaning of asymptotic expansions, we have
$$ T(n) \sim n
- n^{1 - \log_2
\left(\frac{2}{3}\sqrt[3]{17+3\sqrt{33}}
- \frac{4}{3} \frac{1}{\sqrt[3]{17+3\sqrt{33}}} -\frac{2}{3}\right)}.$$
(The exact coefficients on these terms are calculated above.)
