Capacitor multiplier does not remove ripple
The all important component that is missing is in the red box below: -



Taken from the EDN article linked by OP.

You have to have the resistor in the red box to bias the base at a few volts below the incoming supply voltage or it just won't work properly. So if your input supply is 20 volts, arrange for R2 and R1 to provide a DC voltage of about 16 volts at the base. This will means about 15.5 volts at the emitter on light loads and maybe 14 volts on the full load.

The parallel impedance of R1 and R2 should be 220 kohm to acheive the same cut-off as you quoted when C2 is 10 uF.