Besides that, I just googled CMOS output voltage levels, and I had to correct my previous post. The max is apparently defined at 50mV, which is a surprise to me and puts a new twist on this. At first, I didn't think anything of 43mV. 43mV is small. But compared to 50mV max, it is a lot.
If his output pin is low, and it's only connected to a FET gate, then should it even measure 43mV? CMOS output is through a FET, itself, which is a purely resistive voltage drop. And with no load, would it even be that high? Where is this 43mV of drop coming from?
So if an output pin is rated for 25mA, let's say, it has to be able to sink 25mA and still maintain 0.05V, max. At a microamp (to cover gate leakage, lol), it should be much less. At 43mV, if the Atmega actually meets spec, this suggests the pin is sinking at least (if not more than) 4/5th of its max output rating simply holding a signal FET gate down. (Or that his pin maybe isn't doing exactly what he thinks it is. Which sounds more plausible?)
So I have a PIC project already wired up on my bench. And I just measured voltage on a PIC output pin digital low with no load. Supply voltage 5V. Pin measures 0.000V. Nada. (And of course I also verified it gives ~5V when the pin goes high to validate my test setup!)
For giggles, if that 43mV were coming from an intermittent high output or an internal pullup resistor, 0.043V/5V is a duty cycle of 0.86% at 5V (or 1.3% at 3.3V), if I'm not mistaken. Enough to give an LED a decent glow. Due to the capacitance of the gate and the rise/fall times of the pin output FETs, the actual duty cycle of the pin might be considerably smaller than 1%, if the frequency is high. This is a situation that might occur... if the ISR is using that pin, for instance. A UART ISR, for example. Just one possibility to consider.