20mA is usually around the max rated forward current. It's a fairly conventional rule of thumb to select 10mA. That'd give a resistor of 160R, or thereabouts (selecting 10mA as the basis of the resistor calculation gives you a lot of wiggle - you can halve it and still be within the working limits, chances are you could double it and it'd still work). You get less flexibility as you chose a resistor for a larger forward current, of course.

An idealised, simplified model of an LED would presented a voltage source of V-forward (1.7, above), and 2 in series would be 3.4V. That then would mean without a series resistor you're presenting 1.6V (as you say) into a dead short - theoretically infinite current and the blue smoke escapes. Of course, the real model of the LED would have some internal series resistance, and there's the series resistance of the cabling, and the PSU. It might work. It might work for a while. It might work for not very long. But you have spares!, and if you want any resistors thrown in the post I can do that. It's an interesting experiment in any case, and you can but learn from it.

With Mach, can you confirm that with stationary spindle you get 0 RPM. Then at very low RPM you get some RPM (maybe not the right RPM, but some). The first thing you want to make sure is that the encoder input is working before getting too hooked into the inner workings. You should be able to monitor P12 input on diagnostics page and rotate the spindle by hand to find the index position - just confirm that is working with 1 PPR.