Running a stepper motor at under its rated current will affect both its maximum speed and torque (both holding and dynamic), though depending on the driver type, it mostly effects the torque.
Firstly, The stepper motor driver you are linking requires power rails of 7-30V. Its logic interface is 5 or 3.3v.
Furthermore, your stepper motor is 3.6 ohms per winding, so even with 3.3V rails, your motors are going to pull ~900 ma, which will cause power rail brownouts.
Fortunately, in this context, the stepper motor controller IC provides chopped current limiting, which is a technique which limits the maximum amount of current it runs through the stepper motor. Therefore, you can run the stepper motor you have off the wall-wart you have, a the cost of reduced torque.
In a system which uses current chopping, the voltage largely determines the motors maximum speed, and the current the torque. There is interactions, but they are fairly small in effect size (assuming you are not at extremely low voltage or current), proportionate to the effect of changing the relevant characteristic directly.
I urge you to carefully study the A3967 datasheet (the A3967 is the stepper motor IC the linked driver uses).
Furthermore, the designer of that stepper motor driver provides a fairly decent introduction into the concepts of how the device works here, which, from your question, I don't think you have really read. You should carefully read it before you hook anything up, lest you damage one of your parts.
Additionally, the stepper motor driver you link already has a voltage regulator, so you do not even need the voltage regulator you included in your post (SFE sku: PRT-00114).
See the schematic for the Stepper Driver Here. The voltage regulator is IC2.
You need the motor torque constant. It's sometimes called Kt and/or Ke. This parameter will let you relate current to torque. Then you can work backwards, starting with the torque you want, figure out what the current must be, and then ohms law will tell you what voltage will push that current.
Best Answer
No, you have the right rating, it is just that you have a much better motor than in the tutorial.
The motor in the tutorial is a high voltage, low current one. The problem with this is that as a result it has a high inductance, and meaning that it take a while for current to rise to the rated value after each step, which means that the torque when stepping rapidly will be much less than the rated torque. To overcome that, a higher voltage driver (say 36 volts) would be needed with a chopping current regulator.
Your motor with a nameplate spec of 1.68 Amp per phase will be built for a lower steady-state voltage (probably around 3-4 volts, though that is only a guess), and will have a lower inductance. As a result, it will probably perform well when used with a chopping current regulator powered from 12v.
Of course it is possible your current driver is not rated to provide that much current.
Decent 3d printer kits for example may supply low voltage high current motors for the X and Y axis and extruder; but they often cheap out and supply a high voltage low current motor for a screw-drive Z axis, figuring that the Z axis does not need to produce a lot of torque (or if direct drive, does not really need to move very quickly). A yet better kit might use decent motors there, too.