For example, if $$$n = 3$$$, the frog can make $$$7$$$ jumps with the grid

4 3 7
8 0 6
5 1 2

and it can be shown that you cannot do better than this.

Here is a visualization:

The frog stops at the cell labeled $$$8$$$ as such a jump would throw him off the grid.

First, let's handle the $$$n = 1$$$ case, as the frog will make exactly $$$0$$$ jumps, since the number $$$0$$$ would send him off of the grid. So the answer is $$$0$$$ and the grid is also just $$$0$$$.

For $$$n \ge 2$$$, notice that a lot of numbers will immediately throw the frog off the grid, no matter where the frog is currently. Specifically, any number with a jump magnitude greater than $$$n - 1$$$ will always throw the frog off of the grid.

The lower bound for these numbers is $$$4(n - 1)$$$, since $$$\lfloor \frac{4(n - 1)}{4} \rfloor + 1 = n$$$. The number directly below it, $$$4(n - 1) - 1$$$, is the highest number to have a jump magnitude less than or equal to $$$n - 1$$$, and is therefore the highest possible valid jump.

It should also be pointed out that $$$4(n - 1) - 1 = 4n - 5 \le n^2 - 1$$$ for all $$$n \ge 2$$$, so the maximum jump that won't throw the frog off the grid, $$$4n - 5$$$, is always present for grids with $$$n \ge 2$$$.

Therefore, the only numbers that can be included in the frog's path are from $$$0$$$ to $$$4n - 5$$$, inclusive. This is $$$4n - 4$$$ total numbers, giving us a potential of $$$4n - 4$$$ jumps. But can $$$4n - 4$$$ jumps ever be made? Since $$$4n - 4$$$ is a multiple of $$$4$$$, every jump with some magnitude in some direction will always have a counterpart: a jump with the same magnitude in the opposite direction. For example, $$$0$$$'s counterpart is $$$2$$$ and $$$1$$$'s counterpart is $$$3$$$. So, if the frog were to use all $$$4n - 4$$$ jumps, the jumps would cancel out, and he would end up exactly where he started, which would lead to an infinite number of jumps / a cycle.

So, at most, we can use $$$4n - 5$$$ jumps. It turns out that this can always be done for $$$n \ge 2$$$ with the following construction:

Basically, the frog makes jump to the right of magnitude $$$n - 1$$$, then to the left of $$$n - 2$$$, then to the right $$$n - 3$$$, etc down to a jump either left/right (depending on the parity of $$$n$$$) of magnitude $$$1$$$. Now the frog is close to the center column of the grid (slightly to the right for even $$$n$$$).

Then the frog makes a jump upwards of magnitude $$$n - 1$$$, then a jump downwards of magnitude $$$n - 2$$$, then up $$$n - 3$$$, etc down to a jump either upwards or downwards (depending on the parity of $$$n$$$) of magnitude $$$1$$$. Now the frog is close to the center of the grid.

Next, the frog "reverses" the sequence of horizontal jumps, as seen in the image. After doing so, the frog is on the left side of the grid again, and all of the valid horizontal jumps have been used (for each magnitude, either its right jump was used at the bottom row and its left jump at the center row, or vice versa).

Finally, the frog reverses its sequence of vertical jumps, as seen in the image. Except it can't make the final jump downwards with magnitude $$$n - 1$$$, as that would lead it back to its original position. We have used exactly $$$4n - 5$$$ jumps.

...

But wait! For $$$n = 2$$$, this construction would yield the grid

3 2
0 1

which creates a cycle.

For $$$n \ge 3$$$, the construction always works, as you can always put a number with a jump magnitude of at least $$$n$$$ in the top-left corner, and the frog will stay there permanently (this is because, for $$$n \ge 3$$$, $$$4n - 5 \lt n^2 - 1$$$, meaning that the first number with jump magnitude of $$$n$$$, $$$4n - 4$$$, is always present for grids with $$$n \ge 3$$$).

But for $$$n = 2$$$, the answer can't be $$$3$$$, as the final jump would always lead to a cycle. The answer can't be $$$2$$$, either, as attempting to create such a grid would create a cycle:

2 3
0 1

Thus, the answer must be $$$1$$$:

2 1
0 3

The end.

Bonus: can you solve it so the frog jumps the (finite) maximum number of units?