I will focus on just a little bit of one of your questions - the relationship between compressibility, density and pressure - and per my comment, recommend that you narrow down the scope of your question.
As you know, in a gas we experience "pressure" because molecules hit the walls of the containing vessel. When I double the number of molecules in the same volume at the same temperature, I double the number of collisions (each imparting on average the same momentum) and thus double the pressure - this is the familiar ideal gas law.
Now when the size of the molecules becomes a sizable fraction of the volume, the rate of collisions goes up. Imagine a pingpong ball between two walls. If the distance between the walls is large compared to the size of the ball, the time for a round trip is inversely proportional to the size of the ball; but as the distance approaches the size of the ball, the rate of collisions goes up rapidly.
I think a similar thing happens with "nearly incompressible" liquids: there is a small amount of space between the molecules, but they are permanently bumping into each other and into the walls of the vessel. As you increase the pressure, they bounce more frequently as they have less far to travel before they collide with another molecule (or the wall).
All this is still treating the liquid like a non-ideal gas. In reality, not only do you have the finite size of the molecules, but also attractive forces between them. Both these things make the picture a bit more complex than I sketched. But I would say that the above reasoning nonetheless applies (with caveats).
As for the experiment you described with stoppers on the inside or outside - there are other things going on there as you go from the static (no flow) to the dynamic (flow) situation - the water needs to accelerate before it will flow out at a certain velocity. But I think all that should be the subject of another question.
Okay, so you asked specifically about "a cylinder filled with propane gas", and if it's really filled with the gaseous form then that will distribute evenly.
However you probably meant that the cylinder was filled with liquid propane, and that is a lot harder to deal with. If the system could handle the pressures you might try a bunch of successive runs where you try to boil some of the liquid evenly and then condense it evenly within the tank, otherwise you have to wait for natural processes of condensation to eventually spread out the liquid, and there's a good chance that they might not.
There is a nice effect called the siphon effect, it says that if you have two open-air tanks filled with different amounts of water, and you fill a tube completely with water and it arcs over the edges of the tanks to connect them, then if you leave them for a while you will discover that they have the same water level. I do not mean that they will be filled evenly because they might have different shapes and one could be lower or higher than the other; I just mean that if you took the water level from the one as a 2D plane extending across the space outside the tank and eventually intersecting the other tank, you would see that this plane also intersects the other tank. What happens can simply be described as "the water seeks its minimum-energy configuration, some of the water from over here can have a lower energy over there, and because the tube is 100% full of water it does not form a potential energy barrier to this sharing of water."
Now this is a very suggestive result for your purposes, but it comes with a lot of stipulations, too. The tanks are open-air, but why? It's so that I don't need to consider in my energy calculations "okay this tank has more water in it, but also its air is pressurized and that contains some energy too." In your case you would like two tubes connecting the cylinders, one that could equalize pressures while one could allow fluid to flow, and then they could come to the same fluid-levels via siphon. But in order to be very sturdy usually these tanks only have a valve on one side and that's it.
So if you want to do the same thing you probably need to cycle the same valve from "equalizing pressures" to "equalizing water levels" back and forth over-and-over again.
The easiest way to do this would be to stably fix the 5 cylinders side-by-side in a big rack that can rotate slowly, so that when the valves are all upside-down then they share some propane-liquid, and when the valves come right-side-up then they equalize their propane-gas amounts. This will still take a while to equalize all of the propane, but it should eventually do it and it could probably solve the problem over the course of several hours while being left to itself. What you really want to increase the speed is to find a regime where the tube is consistently half-full of fluid, so that it really acts as "two tubes" and allows free flow between the tanks.
Best Answer
Only the depth of the water matters.
$P=\rho gh$, where $\rho$ is the density of the water, $g$ is the acceleration due to gravity, and $h$ is the depth of the water.
No, water will not rise about the height of the surface of the tote
$P=\rho gh$, where $h$ is the difference between the height of the surface of the water and the height of the point where the pressure is measure, such as the outlet of a hose coming from the tank.