Saw there wasn't a topic for this yet, so might as well start it
I've got some maths behind at which point you should theoretically be able to see a LED switch on, as well as some reasons why especially bargain bin LEDs won't ever go so low.
Fundamentally, LEDs convert electrons directly into photons one to one. Obviously not just like that (the QM is a bit more complicated), but the model relationship is one to one. This means that one coulomb (6.24x10^18 electrons) will produce 6.24x10^18 photons. For a red LED, each photon will have an energy of 1.8eV because... the forward voltage is 1.8V, i.e. each electron contains 1.8eV. Without regarding coulomb losses and that kind of stuff, this is exactly how you can regard LEDs and fundamentally this is the reason why they're so efficient. Practically, electricity-to-photons efficiency of some LEDs is in excess of 70% (which is super close to ideal).
This also means you should expect 1 photon per second at 0.00015 fA, regardless of the color. 10fA, your lowest resolution, should theoretically produce 62 400 photons per second.
There are a few very big but(t)s to this, especially when dealing with traditional LED substrates (e.g. gallium arsenide). LED dies are extremely large for the amount of electrons commutated at these currents, so just by virtue of the thin barrier, you're going to get many fA of pure leakage through quantum tunnelling.
Bargain bin LEDs also aren't (generally) monocrystalline, defect-free or even very well cut and doped dies. In fact, only fairly recently has ion implantation (a 'better' way of doping than chemical bath doping) really started to trickle down to low-cost LEDs. This means there are a lot of small areas on the chip with tiny little (fairly high-resistance) shorts in them where there are misplaced or missing doping atoms, which can contribute to pA or fA of additional leakage current. Plus grain boundaries and other crystal defects.
Even when electrons properly recombine, they might not produce a photon, but one or more phonons ('heat particles'). This contributes to the so-called quantum efficiency, the number of photons produced divided by the number of electrons passing the barrier.
Also, photons do not get produced with a definite orientation in these substrates, so you fundamentally lose a bunch of them inside the substrate (because they're produced in-plane and never leave the die). Then a bunch get absorbed or scattered in the encapsulation. The combined effect of die and encapsulation loss is generally about an order of magnitude for old type LEDs (i.e. you get 1/10th of the light out the front compared to what was produced in the bandgap). High-brightness LEDs fix this problem almost perfectly.
There are also a few charge/magnetic effects that i won't get into; some of them are not very well understood either (e.g. LED droop)
A modern substrate, purpose-made (tiny) low-current high-brightness LED die should be capable of producing measurable amounts of light at 1fA. It still has low quantum efficiency (about 10%) and some leakage, but most of the other effects should be so diminished as to not matter.
Also, pro tip: modern CMOS DSLR camera sensors are capable of detecting as little as 2-3 photons per pixel (50% over noise). They're incredibly useful for these kinds of experiments if you don't have the money or time to get a lab grade photon counter.