There is no need to contrapose collisions of molecules and weight of air!
You can take a syringe without a needle, close the muzzle with a finger and pull the piston a little. There will be a vacuum in the syringe under the piston, atmospheric pressure will push the piston back inside. Where this force come from? Atoms and molecules of the air collide with the surface of the piston and push it. The surface of piston "does not know" if there is some atmosphere around the Earth, it only "knows" some molecules and atoms are constantly bombarding it. By the way, there may be no Earth and atmosphere (f.e. inside the space station) - but from the syringe's point of view the situation would be the same: constant bombardment from molecules around and hence the pressure.
Situation is quite similar with liquids: the surface under pressure "knows" only about the molecules it contacts with. Still it is possible to calculate the pressure using formula F = Spg*H. But it would not be some additional component of the pressure!
Mechanism of air pressure is "bombardment" of the surface by molecules.
Rootcause of atmospheric pressure on Earth is the weight of air.
There may be many different ways to calculate the atmospheric pressure: to divide the total weight of the air over the total surface of Earth is one of them. This approach makes it possible to calculate the atmospheric pressure without going into details how exactly molecules collide with the surface, but the mechanism of air pressure remains the same: collisions between molecules.
Assuming this giant ball of water can hold itself together due to cohesion, wouldn't you still feel the pressure from...well, simply the water molecules themselves, moving randomly in all directions?
This is a pretty unrealistic assumption, and showing what would happen should help explain how.
The cohesive forces allow for a surface tension, which can maintain a pressure difference between the sphere of water and the outside. The pressure difference due to surface tension between an inside and outside fluid and gas surface is known as the Laplace pressure. The Laplace pressure for a sphere is given by the equation $$\Delta P = \gamma \frac 2R$$
where $\Delta P$ is the pressure difference between the curved surfaces, $\gamma$ is the surface tension of the liquid, and $R$ is the radius of the sphere. We can assume in the vacuum of space that the external pressure is 0, so the value of $\Delta P$ will represent the total pressure inside the sphere, if we assume only the cohesive forces are acting.
Now if we look at the surface tension of water, $\gamma_{\text{water}}=71.97 \ \frac{\text{mN}}{\text{m}}$ (I'm assuming standard conditions to illustrate the point; but realistically due to reasons below, I don't think you can calculate the actual surface tension of liquid water in the vacuum of space) and the Laplace pressure equation, we can see part of the problem. Let's assume the sphere is 2 m in radius, since that is likely the smallest radius you could even consider it swimming.
$$\Delta P = \frac {2}{2 \ \text{m}} \cdot71.97 \ \frac{\text{mN}}{\text{m}} = 71.97 \frac{\text{mN}}{\text{m}^2}$$
which is only $0.07197 \ \text{Pa}$. Atmospheric pressure is 1.4 million times greater (and it only gets lower with increasing radius unless you consider gravity). So to explain that aspect, if a giant ball of water could keep itself together through cohesion alone, it wouldn't really feel like any pressure at all to swim inside it.
But that probably doesn't solve all of your confusion, which relates to what I mentioned at the beginning. The unrealistic assumption is more that water would remain a liquid in these conditions at all. It cannot hold itself together due to cohesion, as liquid water at these pressures. It will want to change phases, as mentioned in the other answer. This will all depend on the thermodynamic effects of the fluid, not as much the cohesive effects. It should be pretty easy to see that at low pressure, (such as the vacuum of space with minimal cohesive force) you cannot even have a liquid phase of water. see here for an image
Best Answer
From Pascal's law, we know that pressure is isotropic, which means that at a given location in a fluid, it acts equally in all directions. So, at a given location, the horizontal force per unit area acting on a small vertical surface is the same as the vertical force per unit area acting on a small horizontal surface.
Usually, a room is not hermetically sealed, so it is not totally separated from the atmosphere. Any connection between the room and the atmosphere allows the pressure to equalize (by air flowing in or out). As we said above, pressure acts horizontally also, so air can come through a vertical crack just as easily as through a horizontal crack. In a house, there are typically vents in the attic which allow communication with the atmosphere.
If the room were totally hermetically sealed from the atmosphere, then you could impose any air pressure you wanted inside the room. It would not have to match the outside atmospheric pressure. But, the forces on the walls could get pretty large between inside and outside as a result of the pressure difference, and you would have to be pretty careful so that the room didn't implode or explode. When tornadoes occur, the atmospheric pressure outside drops substantially, and people are recommended to open the windows (to allow the pressures to equalize) in order to avoid the windows blowing out (or even worse).