This is because of Friction, Coanda effect followed by Magnus effect.
In the middle, he also showed how the water followed the curve of the ball. This is mainly because of 'Coanda effect'.
From Wikipedia,
The tendency of a jet of fluid emerging from an orifice to follow an adjacent flat or curved surface and to entrain fluid from the surroundings so that a region of lower pressure develops.
Here, the water is trying to entrain fluid from surroundings. When it comes near to a surface, it cannot pull air from surroundings which makes a low-pressure region between surface and jet. This is because, 'when something is emptied from its place, then something should fill the void to keep the balance.'
Firstly, the ball starts to rotate because of the friction between water and ball surface which is just like a Tesla turbine. As the ball is being hit by the water on one side (not centre), it will push the ball to the other side because it comes in the way of the water. Once the ball starts rotating, the fluid following the surface don't adhere to the surface much longer and drift apart tangentially.
This is where the 'Magnus effect' kicks in. The magnus effect creates a force perpendicular to the jet direction. This force pushes back the ball to remain in contact with the jet on one side. So, the weight of the rotating ball is born by the jet completely.
The magnus force is proportional to the speed of rotation of the ball which is proportional to the velocity of the jet of water.
So, the ball levitates in air as long as the jet discharge is kept constant.
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
The question isn't silly. The speed of each molecule in the liquid is much higher than the speed of either the piston or the water shooting out from the nozzle. At room temperature, for water molecules the average is on the order of 500m/s. And yet, the speed of sound in water is three times higher than that, which implies that pressure can propagate in water at that speed. If you tried to make a jet of water molecules faster than the speed of sound, then you would run into difficulties with the nozzle method, though. You would get a shock front, strong heating, possibly ionization... all loss mechanisms that would restrict the velocity that you are really after. The better way would be to make a piston that shoots out a cylinder/piston that shoots out water. That's a two stage rocket... to squirt water.
Picking up on Aaron's comment: he is absolutely correct that one can find nozzle shapes which convert a hot, high pressure gas into a clean supersonic flow. That is the major requirement to building efficient rocket engines. The engine nozzle in that case is a thermodynamic machine which converts the internal thermal energy of the gas into a directed flow that is many times faster than the speed of sound in the gas.