Amps/phase in stepper motor
The worst case total current is the sum of the currents, easily illustrated with unit cancellation:  

  1.5 A/phase * 2 phases  
= 1.5 * 2  A/phase*phase  
= 3 A (total)  


Keep in mind that the current is not always 1.5 A per phase. If each motor phase draws 1.5 A at e.g. 12 V, lowering the voltage by half (to 6 V) will also cause the winding resistance to limit the current to half of its value (0.75 A in this example).  

Also, depending on the switching scheme, the driver might power only one phase at a time while stepping forward, known as "wave stepping" (+A, +B, -A, -B), power both phases all the time, known as "full stepping" (+A+B, -A+B, -A-B, +A-B), or only power both phases at the same time during "full steps" as a combination of the two, known as "half stepping" (+A, +A+B, +B, -A+B, -A, -A-B, -B, +A-B). Full stepping will draw double the current of wave stepping, but it will also produce more torque. Using the above example, the motor would draw (at 12 V and neglible speed) 1.5 A if wave stepping, 3 A if full stepping and alternate between 1.5 A and 3 A if half stepping. Then there is microstepping, which usually draws about the same amount of current regardless of the current fraction of a step due to the phase current being varied as a smooth sine wave.  

Likewise, if the motor is being spun at any significant speed (instead of peing powered just to hold a position), the back EMF induced by the rotating magnets within will induce its own, reverse voltage on the motor phases which effectively lowers the voltage "seen" by the coils, as if you lowered the supply voltage. Thus a stepper motor driven at constant voltage will actually draw less current at high speed, and since torque is proportional to current, the maximum torque drops too, potentially causing skipped steps.

For this reason, stepper motors are often driven with constant current drivers, which continously adjust the phase voltage so that the current (and torque) stays constant.