JohnDoe,
the torque required to drive the mechamism depends on what you attach to the other end of the worm gear. If, lets say, your mechanism is a wheelbarrow, where the wheel is driven by a motor through a worm gear. Say your worm gear has a gear ratio of 100:1, the wheel diameter is 0.5 m, and you want your wheelbarrow to be able to pull a load of 100 kg up a 10 degree slope. That means, friction losses aside, that the force required to push the wheelbarrow up the slope is 100 kg x 9.81 N/kg x sin(10 deg) ~ 171 N. In order to produce that force, the torque driving the wheel needs to be (force times radius) 171 N x 0.25 m ~ 43 Nm. Since the worm gear has a gear ratio of 100:1, the motor will need to produce just 0.43 Nm.
Further, if you want to drive your wheelbarrow up the slope at say 5 km/h = 1.4 m/s, the wheel needs to rotate with a speed of 0.9 revolutions per second or 5.6 radians per second. This means your motor will need to produce 43 Nm x 5.6 rad/s ~ 240 Watts of power, or about 1/3 hp, at 90 revolutions per second or 5400 rpm..
(Edit2: corrected error, again..)