Ohm's law says that if you use a 12V supply, an NDP6020P Mosfet, ULN2803A column drivers and 3V LEDs with the 27 ohm series resistors then the LEDs will blow up with 289mA peak current (18mA average current) unless you re-calculate and replace all the resistors in series with the LEDs.
I can't remember, I think you are using a 5V computer. With 5V G-S the FET can do 20A with 0.8V D-S, 10A=0.4V, 5A=0.2V
I think the part needs 4 to 5 volts of drive. Works at 3v at lower current.
If you want more current then change the resistor. Changing the supply voltage puts more voltage across the resistor which equals more current. Changing the resistor also changes the current.
Ohm's law says that if you use a 12V supply, an NDP6020P Mosfet, ULN2803A column drivers and 3V LEDs with the 27 ohm series resistors then the LEDs will blow up with 289mA peak current (18mA average current) unless you re-calculate and replace all the resistors in series with the LEDs.
Since you've got +12 Volts available, I would use a level shifter to drive a high side N-channel mosfet for the column driver.
In the attached circuit, Q1 can be almost any common NPN small signal transistor. Likewise Q2 can be any common PNP transistor. The mosfet cn\an be almost any n-channel part that can handle the load current.
The collector of Q2 will switch between 0 and +12 Volts. So that will drive the gate of the mosfet 7 volts above the 5 Volts it is switching.