Well, let's do the math.
You've got 8.01 μF of capacitance before the common-mode choke (aside: are you sure a Y5U ceramic capacitor is safe to put across the AC lines? That sounds like a bad idea, it should probably be a class-X capacitor. And why are you using Y5U?), and 5.002 μF after it. Let's round off the smaller capacitors and just say we have 13 μF across the line. I'll also assume the common-mode choke is ideal and has zero differential-mode impedance.
I'll assume you're in a 50 Hz, 230 V country, because that makes the numbers work out to closer to what you're seeing. At 50 Hz, 13 μF has a reactance of -j245 Ω. With 230 V across the lines, that will consume a little under an amp of current. This is entirely reactive power, so nothing should get hot--you're just charging and discharging capacitors.
Considering that capacitors can vary quite a bit in their capacitance, 1.4 A isn't out of the realm of possibility if your capacitors are higher-than-nominal capacitance. I can also see at least one other capacitor connected across the lines in the screenshot you've shown, so there may be more as well. I also don't know what voltage or frequency you're using; I made a guess, but higher frequencies and higher voltages will increase the current.
The 1 MΩ resistor and the varistor won't affect things significantly, as they're both going to be much higher impedance than the capacitors.
