Add two resistors:
From +5V to the transistor collector and between pins 3 and 2.
That would act as a potential divider and set the voltage on pin 3 when the transistor is off (= current enabled).
Changing one or the other resistor value, preferably the one to ground, would change the high voltage on pin 3 and change the LED current.
The internal current source to pin 3 is only 8uA so using external values that pass eg. 0.5 to 1mA or so should override that with little error.
eg. used 4k7 from pin 3 / collector to +5V then 4K7 and lower values between pin 3 and ground.
Or both fixed at 4k7 and add parallel resistors from 3 to ground to reduce the current from the maximum.
For ones that need maximum current you can omit the resistor to ground and leave the one to power in circuit - it would work the same as now.
Or you could use a 4k7 preset wired as a variable resistor between 3 and ground, so you can calibrate each unit without changing components. (Or eg. 3k3 preset and 2k2 fixed, so you can go from around 100% to 40% but not all the way to zero - it depends on what range you need).
With both resistors fitted, the voltage on the set pin is:
5V divided by the sum of both resistors, then multiplied by the value of the one from 3 to ground.
4k7 + 4k7 is exactly half supply, so 2.5V, maximum brightness. The resistor to ground can be left out and give the same result.
4k7 and 4k3 = 5V / 9, * 4.3 = 2.39V
4k7 and 3k9 = 5V / 8.6, * 3.9 = 2.67V
4k7 and 3k3 = 5V / 8, * 3.3 = 2.06V
4k7 and 2k7 = 5V / 7.4, * 2.7 = 1.82V
4k7 and 2k2 = 5V / 6.9, * 2.2 = 1.59V
4k7 and 1k8 = 5V / 6.5, * 1.8 = 1.38V
etc.
You can see the relative current each voltage gives in the chart from the data sheet that Ron posted earlier.