so we have to be careful to specify torque at a certain slip. small motors can sometimes safely be run at the breakdown torque at lower voltages without burning up the motor, but larger motors cannot. so yes, in theory that motor could deliver 750 watts to the shaft.. at the breakdown torque of the induction motor. (because break down torque is usually about twice the motor's rated torque) but in the case of a 1.5KW spindle motor.. that would almost certainly burn it up due to I^2r losses in the coils.
so anyhow:
motor torque or magnet force is proportional to the flux density squared, flux density is set by volts/hz. (this is why neodymium magnets are about 10 times stronger than ferrite magnets.. the flux density they can deliver is about 1T vs .3T for the ferrite)
one of the things you have going for you with small induction motors is they are designed to run saturated, so dropping the voltage does not drop the flux density as quickly, power factor also improves which means less current. but only to a point.. this is why i said "as a general rule..." you can take advantage of the better power factor and less saturation to overload the motor slightly and stay below the thermal limits
so if you've ever seen a 1.25 service factor 230volt motor that says "useable at 208 volts" its because when you run that motor on 208volts the 13 percent reduction of nominal voltage (240/208) uses up the entire .25 service factor.