Probability of three dice falling in the same quadrant of a box

The possible combinations are not equi-probable. For instance it is more probable to have 3 dice in 3 known different quadrants than in a single one. You can not get the probability of a combination by taking the inverse of the number of combinations. So your result $\frac{1}{16}$ is correct.

The first take and the second take are the same. The point is , in take 1, what happens is you are inherently fixing the quadrant $x$ in which you want the dice to fall. In truth the dice could fall in any of the four quadrants, but they all have to fall in the same one. Thus, $\frac{1}{64}*4 =\frac{1}{16}$ is the right answer without doubt.

As for the third answer, you may tell the solution giver: It's quite simple. At the end of the roll, let $x_i$ be the number of dice present in quadrant $i$, $i=1,2,3,4$. In the end $x_1+x_2+x_3+x_4=3$, and each of these numbers $x_i$ is between $0$ and $3$. How many combinations of $x_i$ are possible?$\binom{6}{3}=20$. But the combinations are not equiprobable : in fact they are distributed multinomially.

Assuming equal size of each box (more precisely equal probability of ending up in each of the boxes) the solution is 1/16. Your second take is correct. Alternatively take your first take (which gives the solution for a specific box out of the four boxes) and multiply by four.