Practical values for pullup resistors could be anywhere in the 4.7k to 20k, even 50k. The value is not always very critical. Try one of those, that may do the job.
You are correct, the datasheet is talking about whatever will be driving that OE pin. Now, the driving pin has some maximum current sourcing/sinking capability—its datasheet should specify this value.
Say your driving pin can sink 10mA as a maximum, just like you said. Yes, your calculation is correct in terms of protecting that pin from burning out. If you go any lower than 330 ohms, you're exceeding the sinking current rating.
But in practice, you don't want to be sinking that much current through that pin, that is a maximum rating, not a recommended operating condition. Take for example the datasheet you have attached.
In page 4, it specifies a test current, called \$I_{OL}\$ of 50uA, this is the test current when the voltage is driven low by the internal logic. Ideally this voltage should be zero or very close, when driven low but in practice they tell you, it could sit at a maximum of 0.1V. They do the same test with \$I_{OL}\$ of 8mA, and notice that now maximum voltage that you could see, when driving the pin low, increases to 0.36V.
With that in mind, you may want to try a pullup resistor of \$R_p=\dfrac{3.3\text{V}-0\text{V}}{50\mu A}= 66\text{k}\Omega\$
The problem with decreasing the pullup resistor value, is that more current flows when the pin is driven low and the output voltage increases (as you see with the two test currents in the datasheet) and it could be high enough that it is not a logic 0 anymore. Even if you try the case when \$I_{OL}\$ is 8mA, for this case, the maximum specified output voltage is 0.36V, which is still a logic 0. The resistor value turns out to be 412 Ohms. So 10k, for example should work.
Those numbers given here do not apply to your case because I used the values provided in the buffer datasheet, not your driving Ic, which you said it is a microcontroller. But they do give you an idea of how to calculate the values of a pullup resistor for your specific situation.
There is an upper limit for the value of the pullup resistor, as well, and it has to do with the leakage current going into the pin when the internal transistor driving the pin is 'open'. But I think it isn't necessary to go there.
Of course, if your controller can drive the pin both high and low, you may not need a pullup resistor at all. Hope this helps.
