Have a look at the typical characteristics of a [CD4069UB][1] gate: 

[![enter image description here][2]][2]

As you can see, the gate with no feedback (typically) acts as crude comparator that switches at approximately Vdd/2 (which is 4.5V in this case). 

With feedback it can be considered an amplifier with a transfer function as follows: 

Vpin1 = 4.5V, Vpin1 = (Vo + Vin)/2  => Vo = 9V-Vin


Now we can easily calculate the transition voltage for each LED from top to bottom, since it's a  simple voltage divider (the gate inputs draw negligible current): 

[![enter image description here][3]][3]


The right-hand column indicates the difference to the next LED, which is fairly constant. 

It should be mentioned that this circuit works "typically" and there is no guarantee that any given gate will behave that way, in fact if  you look at the worst-case characteristics: 

[![enter image description here][4]][4]

.. it may not work very well at all. Also both the transistors are partly 'on' at the transitions (and the input gate is biased so it also has both transistors on) so the circuit draws extra mA that are wasted. 

So it's an okay toy, but not really something  you'd want to ship. 


  [1]: https://www.ti.com/lit/ds/symlink/cd4069ub.pdf?ts=1613823343137&ref_url=https%253A%252F%252Fwww.google.com%252F
  [2]: https://i.sstatic.net/1Mpvb.png
  [3]: https://i.sstatic.net/xr4Sk.png
  [4]: https://i.sstatic.net/q9Clw.png