A given section of pipe will have a flow equation like:
$\Delta P = \gamma w^2$ where $\Delta P$ is the loss of head pressure over the length of pipe, $\gamma$ is a coefficient related to length, diameter, Reynold's number etc., and $w$ is the flow rate. You can find expressions for these in engineering books for example.
Now imagine we have a pipe and we want to put one small hole in it. We can analyze it by analogy with an electric circuit, where the hole is represented by a large shunt resistance added at that point going to ground. Let $P_0$ represent the fixed pressure driving the flow (analogous to a voltage source), $P_1$ be the pressure at the point of interest near the hole, and let $\gamma_{1,2,3}$ be the flow coefficients for the section of pipe upstream of the hole, downstream of the hole, and the hole itself. Let $w$ be the flow rate upstream of the hole, $w_2$ downstream, and $w_3$ be the flow through the hole itself. Then you have a system of equations you can solve for $P_1$:
$P_0 - P_1 = \gamma_1 w^2$
$P_1 - 0 = \gamma_2 w_2^2$
$P_1 - 0 = \gamma_3 w_3^2$
$w_2+w_3 = w$ (Assuming incompressible flow)
The case of no hole is given by $\gamma_3 = \infty$. Since the hole is small, you can assume $\gamma_3 >> \gamma_2, \gamma_1$ and derive a perturbative expression for $\Delta P_1$ in terms of $\gamma_3^{-1}$. I will leave this as an exercise for the reader. Anyhow, this will tell you the incremental pressure loss for a single hole.
Of course, I have assumed the flow is driven by a fixed idealized source pressure. In reality the source pressure may be a function of flow rate (analogous to the impedance of a voltage source) which would also effect the result, but which could be added into the above equations in a straightforward manner.
What comes first? Pressure or flow?
Without potential, a pressure differential, there is no flow. It's pressure that needs to be there first. So rather than ask what pressure drop you get, it's better to ask what flow you get from applied pressure.
We can never measure pressure drop across an infinite pipe because we can never reach the infinite point. But just considering a very long pipe, there is allot of resistance. So to move fluid you need lots of inlet pressure over what pressure you have at the outlet. And just increase the pipe a little longer, and more pressure is required. At some point you cannot provide the energy needed - even before you reach infinity. You will have a huge (static) pressure differential but zero flow.
Resistive losses in a pipe only occur when there is velocity. But there is something missing in your model - inertance of the fluid. The potential energy you supply at the inlet must provide sufficient force to first accelerate the fluid, then enough pressure to maintain the flow against the resistance.
Just by turning the question around with regards to cause and effect you see all is consistent with what's expected. Pressure doesn't happen after flow. It was there to begin with, however changes when flow is activated.
Best Answer
None of these terms correspond to the friction loss. The equation you wrote down is, as you say, just the hydrostatic pressure equation, which knows nothing about friction. You find the part of the pressure drop that is due to friction loss by comparing the pressure drop you would see without friction (which would be the pressure drop between points 1 and 2, if there was no flow), with the actual pressure drop you just calculated. You should be able to take it from there. Let us know if you need more help.