The pressure at the exit down section is the atmospheric one. The pressure at the top is also the atmospheric one. There is no contradiction nowhere, BECAUSE you cannot apply hydrostatic pressure equilibrium in the vertical direction. Hydrostatic means static, and your fluid isn't static at all.
Let's call the pipe section A, the atmospheric pressure P_a and we have also the gravity g, the characteristic heigth of the pipe L, the fluid density \rho and the dynamic viscosity \mu. And let's play dimensional analysis. Call the vertical upwards coordinate z, r the radial component, and call the vertical velocity component u.
The equations of motion are:
\nabla\cdot \overline{v}=0
\frac{\partial u}{\partial t} +u\frac{\partial u}{\partial z}=-\frac{1}{\rho}\frac{\partial (P+\rho gz)}{\partial z}+\frac{\nu}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)
The last equation is the z-Momentum equation. Note I could have not written P, because it's vertical gradient is identically zero in this problem. Note also that this problem is inherently unsteady. Let's talk about the order of magnitudes. Firstly, one expects to have as much velocity as the one caused by the gravitational force, so that U\sim \sqrt{gL}. Therefore, one defines the Reynolds Number Re=\sqrt{g/L}A/\nu. Non dimensionalizing u=u/U, z=z/L and t=tU/L the equations become:
\nabla\cdot \overline{v}=0
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial z}=-1+\frac{1}{rRe}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)
For Re>>1, that is, for very wide pipes, one can assume ideal flow, and the z-Momentum eqn at leading order (with errors O(1/Re)) is:
\frac{\partial u}{\partial t}+ u\frac{\partial u}{\partial z}=\frac{\partial u}{\partial t}+1/2\frac{\partial u^2}{\partial z}= -1
which can be integrated from 0 to h (the level of water):
\int_0^h \left\{\frac{\partial u}{\partial t}+1/2 \frac{\partial u^2}{\partial z}+1\right\}dz=h\frac{\partial u}{\partial t}+h=0 which is just a nondimensional Bernouilli equation.
Observe that I have integrated the unsteady term directly, because the acceleration is homogeneous inside the fluid. Note that there is no variation of kinetic energy. Also note that from the equation of Continuity one obtains a relation between u and h:
u=-\frac{dh}{dt}. And a second order linear differential equation arises for h:
\frac{d^2h}{dt^2}+1=0.
Then, h(t)=-t^2/2+h_o. Does not sound to you as the free-falling space law for a particle in a gravity field?. Yeah, the water is falling as a whole, as a rigid solid.
To sum up, the hydrostatic balance \nabla P=\rho \overline{g} does not hold in this system. If there is motion in the direction of the body force, there is no hydrostatic balance.
I am still in good shape, am not?
?? <enter confusion> 
I wish arildno was here.
