You can get a minimum bound from energy balance alone. This is as if the fluid has no viscosity, so the force you have to apply over the distance is only due to the kinetic energy required to expell the fluid.
The diameter of the tube is 1 cm, so the area is .785 cm². That means the plunger travel distance is 25.5 cm = 0.255 m.
The fluid is squeezed down to 200 µm diameter, which is a crossectional area of 31.42x10-9 m². The volume of fluid is 20 ml = 20x10-6 m³
(20x10-6 m³)/(31.42x10-9 m²) = 637 m
That is how far the 200 µm stream has to travel in 20 seconds, for a speed of 31.8 m/s. 20 ml of water has a mass of 20 g, or 0.020 kg. The total kinetic energy that is therefore imparted on the fluid is
½(0.020 kg)(31.8 m/s)² = 10.1 J
Now we can solve for the force required over the plunger travel distance to impart this energy:
(10.1 J)/(0.255 m) = 39.8 N = 8.95 pounds
That's actually a lot more than I expected before working it out. It would be interesting to see how much higher the force is when viscosity of the fluid is taken into account. It might be possible that kinetic energy is actually the dominant effect for something with relatively low viscocity like water. Obviously the force would go way up for something thick and gloppy, probably to the point where a typical syringe couldn't handle the pressure to acheive the expulsion time of 20 seconds.
Hmm, that's a interesting point. Let's see what the pressure is. The area of 0.785 cm² is 0.123 in²
(8.95 pounds)/(0.123 in²) = 73 PSI
Which is the pressure inside the syringe required to expell the fluid just due to the kinetic energy requirement alone.
There is yet another effect at work that makes the minimum required force higher, still without invoking viscosity. The speed won't be the same for every part of the flow thru the narrow tube of the needle. The flow will be laminar, so the outer edges will be slower with the highest speed in the center. The average still needs to be what was calculated above, but the power will be higher because it scales with the square of the speed.
The difference is the same as the ratio between the RMS flow rate and the average flow rate. For example, for a linear profile from edge to center, RMS is 22.5% higher than the average. Of course that's a rather unreasonable profile, but it illustrates the concept. I picked the shape of a half-sine as a close-enough profile. This means the flow speed is 0 at the edges and smootly peaks in the middle. Perhaps someone more familiar with fluid dynamics can tell us what the real profile is, but I expect this will come close enough for the purpose of increases the energy requirement due to the spread of flow speeds.
I was too lazy to do the 2D integrals, so I had the computer do the integrals numerically for me. The RMS of the sine peak profile is 17.9% higher than the average. That means the 10.1 J calculated before needs to be increased by this amount. That comes out to:
Force = 46.9 N = 10.5 pounds
Pressure = 86 PSI
As before, this is without the additional force needed to overcome viscosity of the liquid. The only properties of the liquid this relies on are its density, and that flow thru a 200 µm diameter pipe will be laminar.