Electronic – Diode Drop vs Open Drain Output Tolerance

This circuit will work, but not be efficient, power-wise. If the output is high the LED is on and you'll have 20 mA current through LED and resistor.

But with the output low, the LED is short-circuited and therefore off, but the resistor current will be even higher: 50 mA!, which is not only bad for your wallet and the environment, but also for the output pin; most I/Os will only allow 25 mA.

Place the LED in series with the resistor. Then a high output won't light the LED because the output FET is switched off. A low output will draw the 20 mA through LED, resistor and open-drain output.

Note that this will invert the logic: in your first schematic the LED will be off when the output is active, in the second circuit it will be on with active output.

edit
If the output can't sink enough current you'll need an external transistor to increase that.

If the output is active (low) there will flow a 1.5 mA from +3.3 V through Q2's emitter-base junction, R2 and Q1. That 1.5 mA will allow Q2 to source more than 100 mA for a typical general-purpose transistor. You will only get about 22 mA, though, because R1 won't allow more.

If Q1 is off there won't be any base current through Q1, so the LED will remain off. Q1 will have a small leakage current, and to avoid that this would get amplified by Q2 I added R3. As long as the leakage current is less than 0.7 V / 15 kΩ = 50 µA all the leakage current will flow through R3, so that will ensure Q2 will be completely off.

You're kind of in a gray area. The output should not break down until you get to more than 6.5V and you only need 5.6V for a reliable Vih with 8V supply. Remember that "limiting values" aka "absolute maximum" are not intended to indicate where the part should be operated continuously for reliable operation. The recommended maximum output voltage for normal operation is less than 5.5V. Ignore that at your peril- only you know what the consequences of a failure are. It's also better to drive the 4051 inputs right to the rails otherwise the current draw of the 4051 could go up.

You could use a couple of SOT-23 or smaller dual pre-biased BJTs (transistors) rather than the 74LVC07A and avoid any possible problem. That's what I suggest doing.. the ones I linked above are good for 50V and are quite compact, but there are many others.