Assuming you want to maximize the speed while keeping the current consumption at minimum. It's pretty easy to calculate the maximum and the minimum resistor values. This is explained in this datasheet.
Bus capacitance
First, we need to know the bus capacitance. It is true that you can make a rough estimate it by adding the pin capacitance of the drivers. However, if you are developing a product, you can assemble a prototype with whatever test resistance and calculate a more accurate value.
In the referred datasheet, the following equation is derived:
$$t_r = t_{0.7V_{CC}} - t_{0.3V_{CC}} = 0.8473 \times R_{P} \times C_b$$
You can measure the rise time on your test set up and plug the highest in to the equation:
$$C_b = \frac {t_{r{Max}}}{ 0.8473 \times R_{P_{test}}}$$
With this value, you can also validate if your system meets the maximum bus capacitance allowed by the specification.
Maximum resistance
After you get the capacitance, you can plug the capacitance back in the equation to get the maximum resistor values for your system:
For standard mode:
$$R_p(max) = \frac {1000ns}{ 0.8473 \times C_b}$$
For fast mode:
$$R_p(max) = \frac {300ns}{ 0.8473 \times C_b}$$
For fast mode plus:
$$R_p(max) = \frac {120ns}{ 0.8473 \times C_b}$$
Minimum resistance
This is the easiest calculation, just divide the maximum bus voltage drop by your maximum target current:
$$R_P(min) = \frac{V_{CC}-V_{OL}(min)}{I_{OL}}$$
Note: In the datasheet they specify VOL(max). I believe it is a typo. In any case, this answer provides a safe quick calculation.