This problem lends itself to geometric probability since we can view the interval as a range on an axis.

WLOG, let $$n \geq x \geq y \geq z \geq 0.$$

Then we have that the points $(x, y, z)$ which satisfy this restriction form a tetrahedron.

The height of this tetrahedron is $n,$ and the base has an area of $\dfrac{n^2}{2}.$ This makes the volume $$\dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}.$$

Now we have to apply the restrictions from the problem statement. We need to find the region where $$|x - y|, |x - z|, |y - z| \geq 1.$$

From our ordering condition that we imposed, these inequalities reduce to $$x - y \geq 1 \text{ and } y - z \geq 1.$$

These two restrictions form another tetrahedron as shown below.

Note that in the new tetrahedron, all the dimensions have been reduced by $2.$ This makes the height $n - 2$ and the base $\dfrac{(n - 2)^2}{2}.$

The volume is then $$\dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)$$$$= \dfrac{(n - 2)^3}{6}.$$

The desired probability is then $$\dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}.$$

Plugging in all the answer choices, we get that the smallest value such that this fraction is greater than $\dfrac{1}{2}$ is $10.$

Thus, D is the correct answer.