The assumption of constant velocity throughout the tube arises from the following:
- Assuming constant density
- Conservation of mass
In addition to some other stipulations that should be found in the problem description (like the fact that the tube has constant area), and a few other hydrodynamics simplifying assumptions (like assuming the velocity is constant over the cross sectional area, or an equivalent assumption). One way to argue for the constant velocity assumption is to employ the common $\dot{m}=\rho v A$ equation, which gives flow rate given density, velocity, and area, assuming constant velocity over the cross sectional area. Out of these, area is constant from the basic specifications of the tube, and density can be written to have two dependencies, pressure and temperature, $\rho=\rho(P,T)$. Both the pressure and temperature change over the length of the tube, but it's small enough that if you're working under ordinary "bathtub" like conditions they don't matter.
The fact that $\dot{m}$ is constant over the length of the tube can be established by simply noting that the system is at steady state, and conservation of mass dictates that the mass flow rate going in is the same mass flow rate going out, since the tube has a constant amount of fluid in it.
To do a little bit more overkill, $d\rho/dP$ is a constant that plays a role in determining the speed of sound in the fluid, but is very small for "incompressible" fluids, although incompressibility would mean that value is zero. Temperature can have a significant effect, as $d\rho/dT$ is very much nonzero, but the frictional heating over the pipe just isn't much.
Best Answer
Simply put, only three factors significantly affect flow rate:
$$v \propto \sqrt{\Delta h}$$
Pipe diameter: that relation is a little more complicated but volumetric throughput benefits greatly from using a smooth, large internal diameter hose or pipe.
Pipe length: volumetric throughput $Q$ is inversely proportional to pipe length $L$:
$$Q \propto \frac1L$$
Initial throughput caused by the pump has no effect once flow has been established and the pump has been switched off: then other laws take over that determine flow rate/throughput.
Edit: with regards to point 3.
Darcy-Weisbach equation for laminar flow:
$$\frac{\Delta p}{L}=f_D \frac {\rho}{2}\frac{v^2}{D}$$ With $f_D=\frac{64}{Re}$ and $Re=\frac{vD}{\eta}$, and $Q=\frac{\pi D^2v}{4}$ then with reworking:
$$\frac{\Delta p}{L}=\frac{128\eta Q}{\pi D^4}$$
Bear in mind that $\Delta p$ is the pressure difference between inlet and outlet of the hose, then all other things being equal:
$$\large{Q \propto \frac{1}{L}}$$