This is due to a leak from capacitive coupling. In short, a charged wire might be able to charge a little more if it is close to a conductive material (it "pushes" a little further the electrons of this material). By doing so, a slight current is induced in the live cable, which subsequently generates a magnetic field. If another non-live cable runs in its vicinity, it will catch some of this field and a slight current is now induced in the neutral: the current travels for a few milli seconds.
But because this is AC, then the electrons moving to the right will start moving slightly to the left as soon as the voltage reverses; thus repeating the previous steps, and so on: you AC circuit is "closed". The effect increases when the cables get closer to each other, and/or when they run parallel for a long distance.
AC transformers take advantage of the magnetic field phenomenon : if you look at the circuit of an AC transformer you will realize the coppers of the high tension and low tension loops never physically connect to each other; yet supplying power to the high tension loop subsequently powers the low tension loop.
The electric field effect (letting the electrons move a little further) can be modelled by a capacitor "closing" the loop. Note I put this into brackets because, as per the transformer example, a capacitor only lets AC run through, not DC. Here is an illustration, unfortunately the comments are in French, but the schematics of the circuits are enough to illustrate the purpose. (full article here)
At the top you have a typical situation with two switches controlling the light for example from each end of a corridor: for the length of the corridor, you have two wires running parallel to each other. The schematic is the modelling of this situation.
Then the author of the article goes further: we add a real capacitor, shown in blue. Now, when the circuit is OFF, the leaking current goes through the LED and the capacitor: get a large capacitor and most of the current will prefer to go through it rather than the LED. When the circuit is ON, the capacitor is small enough that it rapidly gets fully charged, which means that the current then can only go through the LED. So choosing the right capacity of the blue capacitor is important: get it too low and you will not absorb the entirety of the leak, get it too high and you will needlessly waste power.
This effect was always there, but it is so small that incandescent bulbs would dissipate it before they became hot enough to produce light. With LEDs, which can turn on with only a few milliwatts, things are different and they do get sufficient power to glow a little.