Will the water in the pipe attached to the bottom of the water tank slow down or speed up when it is emptying? Outside the pipe there is atmospheric pressure as above the surface of the water in the container.
What will the differential equations look like - fluid velocity in the pipe and x(t)? The water in the pipe, until it is emptied, is subjected to a constantly decreasing but non-zero force, and at the same time the mass of water in the pipe is constantly increasing. For simplicity, A1>>>A2, so the velocity of the fluid in water tank compared to the fluid in the pipe is negligible
I guess the equations can somehow be derived from energy transfer, i.e. d Ep/dt = - d Ek/dt, but in the kinetic energy term we have to take into account the time dependence of mass m(t).
We assume there is no friction in pipe. We can assume also that at the beginning all the energy is accumulated in the potential energy in the water tank
P.S. 1 Can we talk about steady flow at all in the "long" time interval? After all, velocity depends on the height of the water in the tank, which changes over time
P.S. 2 How to look at this system?
On the one hand, my intuition tells me that energy is constant at any point in the fluid flow, but on the other hand, it will look as if one part of the system (tank) transfers energy to another part (pipe). Analogous to the oscillating fluid in a tank-pipe-tank system

 A: It's not difficult to obtain the DE relating the height, $y$ of water in the tank (initial height, $y_0$) to the time $t$ from release into the pipe, if we assume that the flow in the pipe is slow enough to be laminar, and for acquisition rate of KE to be negligible compared with work done per second against resistive (viscous)  forces.
In that case we can use Poiseuille's formula for $\Phi$, the volumetric rate of flow of water through the pipe, in terms of the pressure difference, $\rho gy$, between the ends of the water-length, $x$ in the pipe.
As water volume is conserved, for a pipe of inner radius $r$ fed from a tank of inner cross-sectional area $A_1$, we have
$$\pi r^2 (x-x_0)=A_1(y_0-y)$$
Here, $x_0$ is the length of water in the pipe at time $t=0$.
Also,
$$\Phi=\pi r^2\frac{dx}{dt}=\ –A_1\frac{dy}{dt}.$$
You should now be able to set up the DE. I find
$$\frac{dy}{dt}=\ -\frac{\pi r^4 \rho g y}{8A\eta x}\ \ \ \ \ \text{in which}\ \ \ \ \ x=\frac{A(y_0-y)}{\pi r^2}+x_0$$
I can't solve this DE analytically. However it's easy to give a qualitative answer to the question posed in your title: the flow rate will decrease because the pressure gradient across the length of water in the tube will decrease. It does so for two reasons: the pressure difference, $y\rho g$ between its ends decreases, and the length, $x$, itself increases!
Why don't we put $x_0=0$? In other words, why don't we assume that there is no water in the tube at $t=0$? In that case, according to our equation, $\frac{dy}{dt}$ and $\frac{dx}{dt}$ would be infinite at $t=0$. Poiseuille's equation wouldn't be applicable at $t=0$ and just after as the infinite flow-rate that it predicts is unphysical.
On the other hand, putting $x_0=0$ would make our equation much neater, namely:
$$\frac{dy}{dt}=\ -\frac{v_1 y}{y_0-y}\ \ \ \ \ \text{in which}\ \ \ \ \ v_1=\frac{\pi^2 r^6 \rho g)}{8A^2 \eta}$$
This equation won't hold immediately after the water has been released into the tube.
