Determine the largest machines and loads, and the largest pair or trio that will run under load simultaneously, and multiply the nameplate FLA figure by 120V or 240V, as applicable.
Also determine the wattage of lighting, heaters, etc. that will also run concurrently, or multiply rated current by 120V or 240V (also as applicable). You want to end up with watts (lighting, small tools, and resistive loads like heaters) and volt-amps (motors, welders, other inductive machines).
Add it all up, and divide by 240V. That's the worst-case average (across both hot legs) load in current, and you should size your feeder to at least 125% of that. More if you think you'll be adding something very large (or a number of smaller things) that will increase the worst-case loading on the feeder, or if it's not likely you'll be able to balance the 120V loads across the two poles of the panel (like big 120V machines).
Remember that 120V loads only load one pole in that new subpanel, but there are two of them (hence the 2-pole feeder breaker). A 60A subpanel ampacity will give you 120A at 120V, or 60A at 240V, or any combination of the two that doesn't exceed either 14,400 volt-amps total (or watts, for resistive loads) and 60A on either hot leg. Balance the 120V circuits on the two hot legs, especially the continuous loads (lighting is considered continuous) to keep current loading even. 240V loads are balanced by definition, since they use both hot legs (what comes out of one goes back through the other, reversing 120 times per second with 60Hz power).
You cannot just add up the current of all the loads, as 240V loads use both poles, and 120V loads use only one, and that has to be taken into account.
I think you'll find 60A is plenty. Unless you're running very large machines under load simultaneously, or kilns or furnaces, or electric heat, or air-arcing (cutting steel with an arc welder outputting very high current and compressed air to blow the molten metal away).
