short answer is take the AC voltage and multiply it by 1.414 (if you want to do a quick calculation, 1.4 will work), for the rectified and filtered voltage. so a quick calculation of 25x1.4=35Vdc for the rectified voltage is close enough. output power calculations can be ballparked as well. with rail voltages of +/-35V, the "ballpark" output power would be 35x0.707 (quick calculations use 0.7) then square the result, and divide by the load resistance, so 35x0.7=24.5, then square it, giving 600.25, divided by the load resistance (8 ohms for example), gives 75 watts into an 8 ohm load. with the 18V transformer:
18x1.414=25.452Vdc
25.452x0.707=18Vrms (seems like we were here before? you're right)
18^2=324
324/8=40.5W into 8 ohms
of course the actual power output will be a little bit less, since no amp actually has a rail-to-rail output, but clipping generally occurs between 0.7 and 1.5V shy of the rail voltages.
you will also need to know the current required for 6 amplifiers delivering full output power into their loads, so for 8 ohms, and each amp delivering 18V into it's load, the current per amplifier is 18/8=2.25A, multiply by the number of amplifiers 2.25x6=13.5A, and add 20% for operating current of various subcircuits and other "overhead" 13.5+20%=16.2A required from the power supply