Simple - assume no voltage at the LED (false, but safe) choose a resistor that will run less than the rated current of the LED at 32 volts.
For an indicator, "Much less" is often fine. i.e. 20mA is almost always safe and visible. So 1600 ohms, more or less would probably be fine. (32 V / 0.020A) Check your actual part for real numbers and try to stay away from running at the limits if you want long life.
Now, at 32 Volts you'd (again, making the safe but wrong assumption of 0V across the LED) be burning 0.64W (32V X 0.020 A) in that resistor, so you'd need a 1W resistor to be safe. You could either reduce the current further, which is fine so long as it's "bright enough" to your taste, or you could use 4 800 Ohm resistors in series, where 1/4W resistors would be happy.
Per Dave Tweed's comments you might want to change that to 5 mA, 6400 Ohms (in my safe assumption model), and less than 1/4 W (but more than 1/8W)
Either way it should have minimal impact on the supply output current, which you estimate as 1000 mA From which we are stealing 20, 5 or "less than 5" if you find that 10KΩ or 20KΩ gives you enough light. Also as commented, this needs to be across the supply, not as drawn.
When running from a lower voltage supply (32 is a lot of headroom) it becomes more of an error to assume 0V across the LED, and you start to need to look at its data sheet and subtract its typical (or minimum) operating voltage from the supply voltage to get the voltage to compute the resistors for this purpose. Or, you may change to a constant-current drive scheme - but neither is really needed in this example, as the error will be fairly minor for any typical LED.