How stable is the 3.3V pin, can you power a sensor out of it?

The 3.3V pin is usually the output of a 3.3V linear regulator. That regulator could be a separate chip on the Arduino board, or, as is the case with the Nano clones, inside the CH340G USB adapter. Either way, it's going to produce a more stable output than the Zener circuit shown in your diagram. The problem is that the regulator will only handle so much current, and can overheat pretty easily. You would need to figure out how much current the 3.3V pin will have to supply. If it's under 100mA, you're probably ok. (Some here would suggest a lower number.) But even then you'd have to see how warm it gets. Above that, you would be pushing your luck.

Zener "regulators" are generally thought to be not very good regulators since the voltage will vary with the current, which is exactly what a regulator is supposed to prevent. They also waste a lot of power because the current draw is the same regardless of the load. What isn't used in the load gets shunted to ground through the Zener. So, oddly, the Zener is at risk of overheating when the load current is lowest.

If the 3.3V pin can't supply enough current for you, I would suggest you look at adding your own linear regulator. It would just be the regulator and two capacitors, versus the Zener and the resistor, so only one additional part. Something like the MCP1700 or 1702 might be a good choice. They come in a TO-92 package.

If space or parts count is critical, and you enjoy living dangerously, and if the devices being powered from 3.3V are always turned on, you could supply most of the current they need directly from the 5V source through a resistor, enough to bring the voltage up to, say, 2.8V under load, and have the 3.3V pin just top things off and bring it on up to 3.3V. I hope I don't have to point out how dangerous this is. But it does work. I had a Commodore computer that was powered exactly that way. Most of the heat is dissipated in the resistor, but the output is still a regulated 3.3V.