You can replace the pull-up resistor with a smaller resistor in series with an LED and then when the Fail pin is asserted (pulled to ground) there will be a 3.3V drop across the LED and resistor. You will have to select a resistor to allow for the LED current to flow. For example an LED with a 2.1V threshold the voltage dropped across the resistor will be 3.3 - 2.1 = 1.2Vmax, and so to achieve a 20mA current you can select a resistor of 60ohm maximum.