How to calculate the voltage across the load of this circuit
EDIT: Your circuit has a dangerous flaw. The peak voltage of a 220VAC source is over 300V, far higher than the rating of the capacitor in your circuit. I would remove this device from your wall outlet immediately and not trust schematics you find on the internet.

Capacitors exhibit a property called reactance. It's measured in Ohms (\$\Omega\$), so it's mathematically equivalent to resistance. In an AC circuit, a capacitor "looks like" a resistor that varies with frequency. The faster the AC frequency, the lower the reactance of the capacitor. At 0Hz, or DC, the reactance is infinite, which is exactly what you'd expect if you put a capacitor in series with a DC voltage source.

The equation for the reactance of a capacitor is:
$$X_{cap}=\frac{1}{2*\pi*f*C}$$
For the circuit you presented, the reactance would be:
$$X_{cap}=\frac{1}{2*\pi*50*1uF}=3183\Omega$$

Now that we have the reactance of the capacitor, it can be modeled as a resistor in the circuit. Applying Ohm's Law, we can solve for current:
$$I=\frac{V}{R}=\frac{V}{X_{cap}}=\frac{220}{3183}=69mA$$
Note that the relevant value here is current, not voltage. This type of circuit, known as a capacitive power supply, acts as a current source, not a voltage source. That means the voltage downstream of the capacitor will vary as needed to maintain approximately 69mA. For your circuit, the voltage will change based on whether the switch is closed or open (the LEDs on or off). It will also change based on the charge state of the battery. 69mA will always be shared between the battery and the LEDs (or the just the battery if the switch is open).

If you want to think of this circuit in a more general form, replace the battery and LEDs with a 1\$\Omega\$ resistor. Now we can use Ohm's Law again to calculate the voltage across that resistor with 69mA going through it. The voltage would be 69mV.

A few things to note:  


Since you have no smoothing capacitor downstream of the rectifier, the current will be AC. The calculation of 69mA is actually the average current of the rectified sinusoidal waveform. It will peak between zero Amps and \$\frac{220V*\sqrt{2}}{3183\Omega}\$ Amps.
69mA is the ideal supply current assuming there is no load resistance or line losses. The battery and LEDs have internal resistance which sum with the capacitor's reactance when applying Ohm's Law. For example, if the total resistance of the load was 100\$\Omega\$, the calculation would be \$I=\frac{220}{3183+100}=67mA\$ instead of 69mA.  
The large resistor across the capacitor can be ignored for these calculations. It exists to safely discharge the capacitor when the circuit is unplugged from the wall outlet. While the circuit is plugged in, it has a negligible effect.