I have learned a lot from all the answers, and it may be that what my point here is so obvious that it doesn't bear mentioning, but it was not obvious to me and the same may be true for some others.
It seems to me that the effect of the grounding (earthing) paths has not been attributed as the cause of the damaging voltage rise on one leg due to a lost neutral.
Evidently in the US we use the TN-C-S grounding system. See https://en.wikipedia.org/wiki/Earthing_system for which a broken neutral is a major safety risk.
If the neutral is lost (completely or partially) the only or main return current path from the panel (ground rods) to the transformer (ground rod) is through the earth, a path which has significant resistance (unlike an intact neutral which has effectively zero resistance).
For the 125 V power draws in the house, the current flow back to the transformer is the difference between the current in the two hot legs. An imbalance in the two legs will appear as a non-zero current in the return path. If the low resistance neutral is lost, this current will cause a voltage difference (V = IR) between the ground rods at the consumer panel and at the transformer. This voltage difference will be subtracted from the voltage of one leg in the house (the higher loaded one), but added to the voltage at the other leg (the lower loaded leg). Hence any equipment on the lower loaded leg will get higher than half the voltage difference between the legs. And there could be a cascading failure because every time a load disappears (as equipment fails) on the higher voltage leg the voltage gets that much higher.
I played around with the infinite grid of resistors model of the path from the house ground rod to the ground rod at the transformer pole and got stumped and humiliated pretty fast. @Harper referred to this in his comment.
Googled it and found a nifty answer https://www.mathpages.com/home/kmath669/kmath669.htm(The answer for the problem set in the cartoon in @Harper 's comment would be -0.5 + 4/pi = 0.773 Ohm.)
This mathpage analysis gives the formula for the resistance between two points on a diagonal separated by m diagonal steps as:
Rmm = R(2/pi)(1 + 1/3 + 1/5 + 1/7 + . . . + 1/(2m-1))
I guess one could estimate the resistance per foot of the soil and then the number of ft to the pole would be m. But my takeaway is there is significant resistance between the ground rod of the house and that of the transformer pole.
I have two ground rods in series and I could disconnect the outer one from my panel and using jumper cables and an extension cord measure the resistance. Not sure I'll be able to motivate myself to do this though. Does anybody know the resistance through 30 ft of "soil" (now very wet Dallas soil)?
EDIT2 I now realize that I would have to disconnect both ground rods to get an accurate measurement and I am not willing to do that. In my yard away from the house I could stab two scrap pieces of cut-off grounding rod and see what resistance I get between them.
I did go out and pound two 18" long pieces of grounding rod 1 ft into the ground 30 ft apart in our very wet backyard. I used a 50' extension cord as an extension of the test leads of my new Fluke 115 true rms multmeter in the resistance mode. Of course, this is a DC measurement and quantitatively meaningless but I'm just reporting what I got. Someone here must know what a valid result should be.
The first value that appeared on the display was ~ 40 ohm and this rose over ~10 seconds to ~ 120 ohm. I can see that a DC ohmmeter is not going to give results meaningful for 60 hz ac, but I'm just reporting what I got. I would guess ~ 2 ohm to 20 ohm impedance for 60 hz.
The resistance along a diagonal in a 2-D infinite grid of resistors R was referenced above
Rmm = R(2/pi)(1 + 1/3 + 1/5 + 1/7 + . . . + 1/(2m-1)).
The sum of the reciprocals of the odd integers (aka the odd harmonic series) does not converge as m increases to larger and larger m. For m > 5 and progressively better for m > 10 the sum of this series asymptotically approaches a logarithmic function
gamma/2 + Ln(2) + (1/2)Ln(m), where gamma is the Euler (or Euler-Mascheroni) constant ~ 0.57722, so
0.57722 / 2 + 0.69315 + (1/2)Ln(m) = 0.98176 + (1/2)Ln(m).
Testing this for m = 7
The sum gives: 1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 = 1.9551
The logarithmic formula gives 0.98176 + 0.5 Ln(7) = 0.98176 + 0.97296 = 1.9547, and the logarithmic formula gets closer and closer as m >.
So the resistance along a diagonal of m diagonal steps is approximated by
Rmm = ~ R/pi (1.9635 + Ln(m)) where m would be the number of diagonal steps between the two nodes.
So we can see that the resistance between the ground rod of a house and the ground rod of the transformer is a logarithmic increasing function of the distance. This means it is a very slowly increasing function of the distance.