There's no single answer to this. That's why NXP doesn't state it explicitly in the datasheet.
Maximum output current depends on several factors:
- IO drive strength (DSE) with value range: 001,010,011,100,101,110,111
- Operating voltage
- Slew rate, fast or slow
The easiest way is to use GPIO output buffer avg impedance from the datasheet to calculate maximum possible current the pin could drive (See Warning below).
See the table below, let's assume you're working on 3.3v with max drive strength (111). So you have maximum current = 3.3v/23 Ohm = 143.5mA. But this is just a coarse value.
The better way to know it is to read the I-V Curve from their IBIS Model. The curve below is taken from the RT1060 series IBIS Model. ~92mA max current @ 3.3v.
just because you can doesn't mean you should. Now, although you can drive 90mA, the real question is for how long? some IBIS Model includes data outside its absolute rating. So, maybe you could drive 90mA, but only, for example, 30 nanoseconds.
Also, you may drive 90mA longer, but you need to keep the chip cool enough so it won't fry itself. In this case, max current is correlated with your ability to keep the Junction Temperature low.
Finally, the general recommendation from NXP is this: use buffer (transistor, etc) to drive your load. Don't drive high power load with GPIO pins.
i.MX RT1062 is a Crossover Processor from NXP, meaning that this is actually a microprocessor equipped with peripherals, not an MCU with super-high performance. Being a microprocessor, the Crossover processor doesn't have the strength like MCUs have.