Mains Power indicator maximum current draw
I'll assume the frequency is 50Hz,as the link says. You have:

\$ (2 \cdot 120 \Omega)+(470k \Omega || 470nF)+(2 \cdot V_{FWD})+(5 \cdot LED) \$.

Let's say the LEDs have \$ V_{FWD}=2V \$

\$ X_C=\frac{1}{2 \pi f C} = \frac{1Meg}{2 \cdot \pi \cdot 50 \cdot 0.47} = 6.77k \Omega \$

\$ Z_{eq} = 470k \Omega || 6.77k \Omega = 6.67 k \Omega \$

The total (RMS) current will be:

\$ I_{total} = \frac{ V_{in} - 2 V_{diode} - 5 V_{LED} }{2 \cdot 120 \Omega + Z_{eq}} = \frac{230-1.4-10}{240+6670} = 31.63 mA \$

Still, due to the large capacitor's reactance compared to the total resistance, you will have a large displacement factor, plus islanding due to the forward drop voltage. The displacement will be (approximately, it doesn't include diode's/LED's resistance):

\$ Z_{tot} = 2 \cdot 120 \Omega + 6.67 k \Omega = 6.91 k \Omega \$

The two 120\$ \Omega \$ are too small so we can leave them aside, therefore leaving us with:

\$ \phi = arctan \frac{R}{X_C} = arctan \frac{470 k\Omega}{6.91 k\Omega} \approx 89^{\circ} \$

Which is almost pure reactive, therefore the power factor will be (considering the simplifications we made) \$ \approx \$ 1.5%.

A quick simulation with the following schematic:



with the following results:



Which are quite close to the calculations, save the power factor who counts the distortions and harmonics, as well.