I think you don't actually want to know the load but rather the hinge moment of control surfaces. The actuator load is the hinge moment divided by the length of the control horn. Below is a rather poor sketch for a typical aileron linkage, but the principle is correct (source):
It already shows one popular way of reducing control forces: The tab, a little auxiliary control surface which moves against the "real" control surface. This reduces effectivity a bit, but forces by a lot. Here, the amount of deflection of the tab is controlled by a spring in its linkage, which is a clever way to adjust its deflection such that the actuation forces become more constant over speed.
Another way of reducing control forces is a horn: An extension of the surface forward of its hinge line, so the aerodynamic loads here balance those on the surface behind the hinge line. The picture below shows the left aileron of the ATR-72 which is moved by mechanical linkage (source)
This way, the lift loads on the control surface are mostly carried by the hinge and only the actuation loads need to be carried by the control rod or actuator. If you think you don't need all those nifty tricks, your actuator and hydraulic system will become much heavier than needed.
Why are two different methods used? The tab reduces the loads from deflection changes while the horn reduces those from angle of attack changes, too. When sized properly, both together will drive the hinge moment close to zero.
Why do I explain all this? It shows that your question does not have a simple answer. Rather, you need to specify exactly how your control surface looks and is moved, and only then can you start to calculate the actuator loads. I also want to show that a subsonic airplane for 10 - 20 passengers will be perfectly flyable with manual controls. The ATR-72 needs hydraulics only for the flaps, the spoilers, the brakes and the landing gear. Avoiding hydraulics for primary flight controls also lets it get away with single redundancy in its hydraulics system.