A nice riddle. With a bit of fiddling you can find optimal solutions for some of the smaller numbers and then combine the rest from there.
The minimal difference is 15 and 4 points difference for 1 or 2 races respectively and then always 1 point if the number is not divisible by 12 and 0 points otherwise.
n=1: Not much choice, as the difference between first and last place is always 15.
n=2: The best you can do is indeed reverse the order with a minimum of 15-11=4. Proof of optimality: One will have at least 15 points, which means the average for the others is (2*73-15) /11 = 11.909..., so at least one of the others needs to have 11 or less.
n=3: The optimum is 1 (it can't be 0, as the total number of points is not divisible by 12, the same is true for every non-multiple of 12 races): 1st+9th+12th place give you 18 points 2nd+7th+11th give you 18 points, 3rd+5th+10th give you 19 points and 4th+6th+8th give you 18 points. So if three drivers each get each place in these four groups once, they will all end up with either 18 or 19 points.
n=4: Optimum is 1. Same procedure with 1st+6th+9th+12th, 2nd+5th+7th+11th and 3rd+4th+8th+10th place for 24, 25 and 24 points respectively.
n=5: The last non-reducible one and the most tricky. Again you can reach 1, but because the number of drivers is not divisible by 5, it is far less neat. First, give 2nd+4th+7th+8th+10th once each to a group of five for 31 points each. Then distribute the remaining places/points among the remaining drivers in the following way (that part I brute-forced with a short program, but considering how fast it was and how many solutions there are, this could have been done by hand as well): Give 15+15+0+0+0, 10+10+6+3+1, 7+7+7+6+3, 6+3+10+1+10, 3+1+1+10+15, 1+0+15+7+7 and 0+6+3+15+6 points to the drivers for a total of 30 each.
n=6,9: Repeat the procedure for n=3 twice or thrice, but change which drivers get 19 points.
n=7: Apply the procedure for n=3 and n=4, again so that no one gets 19+25 points.
n=8: Same with the n=4 procedure twice.
n=10: two times n=3, then once n=4, again possible without giving anyone the extra point twice.
n=11: same with 2x n=4 + n=3
n=12: Optimum is 0: Just give every place to everyone precisely once.
n=13: 3x n=3+ n=4, if you split it correctly, one driver gets 2 of the extra points and all other get 1.
n=14: 2x n=3+ 2x n=4, similar, with 2 drivers getting 2 extra points and everyone else getting 1.
n>14: Repeat n=12 until you reach 14 or less races left with everyone at the same number of points, then use the corresponding solution.