What's the maximum CAN bus frame (message) rate at 125 kbit/s?

Per Olin Lathrop's suggestion, I'll expand on bit-stuffing.

CAN uses NRZ coding, and is therefor not happy with long runs of ones or zeroes (It loses track of where the clock edges ought to be). It solves this potential problem by bit-stuffing. When transmitting, if it encounters a run of 5 successive ones or zeros it inserts a bit of the other polarity, and when receiving, if it encounters 5 successive ones or zeroes it ignores the subsequent bit (unless the bit is the same as the previous bits, in which case it issues an error flag).

If you are sending all zeroes or all ones for your test data, a string of 64 identical bits will result in the insertion of 12 stuffed bits. This will increase total frame length to 140 bits, with a best-case frame rate of 874 frames / sec. If the data bits are the same as the MSB of the CRC, you'll get another stuffed bit there, and the frame rate drops to 868 frames/ sec. If the CRC has long runs of ones or zeroes, that will reduce the frame rate even further. The same consideration applies to your identifiers.

A total of 16 stuffed bits will produce an ideal frame rate of 850.3 frames/sec, so you ought to consider it. A quick test would be to use test data with alternating bits, and see what happens to your frame rate.

The smallest 2.0a (standard) frame you can build is 47bits ...The smallest 2.0b (extended) frame you can build is 67bits ...That INcludes 3bits of inter-frame spacing, and EXcludes bit stuffing ...In theory we can build a frame which will never stuff; In reality, bit stuffing is going to happen quite a lot!

The maximum baud for CANBus 2.0a/b is 1Mbit.
At 1Mb/S, a single (dominant/recessive) bit is 1uS long, ie. 0.000'001 S
So a 67bit frame [the smallest theoretical 2.0b frame] will take 67uS to transmit - before another (67bit) frame may be transmitted.
1'000'000 / 67 gives 14,925 complete frames (+ 25bits of the next frame)

As you are running at 1/8th of that speed, you can get at most 1/8th of the packets
14'925 / 8 = 1'865 frames/Second @125Kb

By the time you are using all 64bits (8bytes) of data, and ASSUMING you have not triggered bit stuffing "errors" by having strings of consecutive 1's or 0's
1'000'000 / (67 + 64) = 7'633
7'633 / 8 = 954

And that's also assuming your wiring is perfect. Is your can bus made from 120ohm UTP cable and capacitively decoupled at both ends? Or some random wire with a 120ohm resistor across one end?

Overall I would say you're doing pretty well to get 90% of theoretical maximum throughput.