Maximum required order for the Boys function
I think your logic seems fine.

When you make your look up tables, you want to make a table for each $n$ up to some value of $n_\text{max}$. Consider you want to calculate an integral that only requires $n=2$, then it would be very inefficient to calculate $F_{12}$ and then down recur.

In your equation (9.8.12) we can notice that in the sum we have the Boys function as $F_{n+k}$. According to "Molecular Electronic Structure Theory", we might need the first $6$ terms of the sum. I.e. when doing your tabulation you will need to calculate all $n$ up to $n_\text{max}+6$.

Also when you do your tabulation you need a maximum range of your tabulated Boys function, i.e. a maximum value for where you can start using the long range approximation afterwards (else your tables will be HUGE). The long range approximation is given as:

$$F_n(x)\leq\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi}{x^{2n+1}}}\tag{9.8.10}$$

For which values of $x$ to start using this approximation for different $n$ you can take a look at Table II in Efficient recursive computation of molecular integrals over Cartesian
Gaussian functions. I am sure newer litterature on this topic exist, but I haven't encountered it.

As a side note, you can also calculate the Boys function with out tabulation (this might be less effective). By using the following formula:

$$F_n(x)=\exp(-x)\sum_{i=0}^\infty\frac{(2n-1)!!(2x)^i}{(2n+2i+1)!!}$$

This formula makes an implementation of the Boys functions much easier. If you are making an integral code from scratch, I would suggest to implement the above formula. You can always speed up the code later. Note with the above formula you can also just calculate the Boys function for the maximum $n$ you need and then use downwards recursion. If your goal is to implement the Boys function via. tabulation, then you can also use the above formula to generate the values for your look-up tables.