Just because the transformer is rated at 10 amps that does not mean the solenoid striker actually draws 10 amps. It probably draws much less. There is not enough information to say what size pot would cut the power in half, for instance. You would need to measure the current in normal operation to do that. But I can give you some guidelines on picking a resistor.
If you want to substantially lower the striking force you will have a problem. The problem is that it takes a certain minimum current to overcome the spring tension and get the striker to hit the chime at all. If the power were cut in half it might not even reach the chime. And if you get it finely adjusted to just barely hit the chime then the least change in supply voltage or temperature or something could push it beyond the threshold and it won't hit at all.
The next thing to consider is the power level. If you are lowering the power in the solenoid then some power will be dissipated in the resistor (or pot) that you mentioned. If that resistor is not rated for that power then it might burn out. This is especially a concern with pots that are not normally associated with much power dissipation. You might be better off with trial-and-error and fixed resistors. There is one thing that you might have going in your favor - the needs of a door chime are very intermittent. Whatever resistor you put in series with the solenoid is not going to have to carry current for very long. It is possible for a resistor to dissipate more power for a very short time than for a longer time. However if it is possible for a visitor to press and hold the door button and if that causes the solenoid to be engaged as long as the button is held then it is possible for a marginally adequate resistor to burn out.
If you have any resistors in your junk box then I suggest trying them out in some methodical sequence. If the chime does not move at all, use a lower resistance. If the chime moves too much then use a higher resistance. This will be cheaper than buying a power-rated pot. Just as a rough guess I will say that your solenoid draws 2 amps. So if you want 2 amps to drop the voltage in half, then you need a 8/2 = 4 Ohm resistor. So it looks like you will be using some very low resistance values. As for power, 8 volts times 2 amps is 16 watts. That's pretty hefty. But if you have a 1 watt resistor it will probable be OK for very short tests.