Calculating the torque value needed to turn an alternator from an electric motor?
First, consider that 1 HP = 745.7W. If your alternator demand is for example, 2kW, then you need 2.7 HP from the motor to drive it.
The details of RPM and torque vs. power you'd have to get from the alternator supplier. Typical alternators were 50-70A at 12V, or about 720W (roughly 1HP.) Bigger ones are more common now as power demand for electronics and other vehicle systems increases. A quick look yields 90-120A for cars, and 200A or more for trucks (especially diesels.)
To get a rough idea of torque, consider a typical engine at a cruising speed of 2000 RPM. The alternator has a smaller pulley, let's assume 1/2 the size of the crank, so 4000 RPM. Let's also assume it's a big alternator (2kW.)
Note that:

Power (HP) = Torque (lb-in) * Speed (RPM) / 63,025

And

Torque (lb-in) = Power (HP) / RPM * 63,025

Using 2.7 HP, working backwards....

Torque = 2.7HP / 4000RPM * 63,025 = 42.5 lb-in, or 3.55 lb-ft.

Want that in metric? Sure:

Torque (N-m) = 9.5488 x Power (W) / Speed (RPM)

So:

Torque (N-m) = 9.5488 x 2000W / 4000RPM = 4.77 N-m

Upsize that somewhat to compensate for efficiency, and this should give you an idea of the size of motor you need.
As far as amps and volts to the motor, that depends on the power supply you have available. But ultimately your supply will need to be (max power out) / efficiency = max power in.
Say your system is 90% efficient. Then you need for 2kW out:

2kW out / 0.9 efficiency = 2.22kW in


Why bother with funky old HP units? Many motors are rated that way. In this case, if your goal was to use line voltage to power an induction motor, you need roughly a 3HP motor to guarantee 2kW out at the alternator. That's a pretty big motor, which would need a 30A / 117V feed or 15A / 230V (like a dryer plug, say.)
If your limit is a normal 117V/15A (20A breaker) outlet, you'd be limited to 2HP in that case, or 1500W.