The face diagonal of a cube has length $4$.  Find the value of $n$ given that $n\sqrt2$ is the $\textit{volume}$ of the cube.
1. Let the side length of the cube be \( s \). The face diagonal of a cube is the diagonal of one of its square faces. For a square with side length \( s \), the length of the diagonal is given by \( s\sqrt{2} \).
2. According to the problem, the face diagonal is \( 4 \). Therefore, we have:
   \[
   s\sqrt{2} = 4
   \]
3. Solving for \( s \), we divide both sides by \( \sqrt{2} \):
   \[
   s = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}
   \]
4. The volume \( V \) of a cube with side length \( s \) is given by \( s^3 \). Substituting \( s = 2\sqrt{2} \):
   \[
   V = (2\sqrt{2})^3 = 2^3 \cdot (\sqrt{2})^3 = 8 \cdot 2\sqrt{2} = 16\sqrt{2}
   \]
5. The volume of the cube is given as \( n\sqrt{2} \). Therefore, we equate:
   \[
   n\sqrt{2} = 16\sqrt{2}
   \]
6. Solving for \( n \), we divide both sides by \( \sqrt{2} \):
   \[
   n = 16
   \]

The final answer is \( \boxed{16} \).