Let the cube be $[0,1]^n$. I'm assuming you mean the cross-section perpendicular to the longest diagonal, from the all-zeros corner to the all-ones corner. This is the cross-section corresponding to the plane $x_1 + \cdots + x_n = n/2$.

Alternatively, then, you want the probability density function of the sum $S_n$, of $n$ independent uniform(0,1) random variables, evaluated at $n/2$.

By the central limit theorem, the $S_n$ have asymptotic normal distribution with mean $n/2$ and variance $n/12$. So the answer should behave asymptotically like $1/\sqrt{2\pi n/12}$, i. e. like $\sqrt{\pi/6n}$.

It appears that this distribution is called the Irwin-Hall distribution.

Let the cube be $[0,1]^n$. I'm assuming you mean the cross-section perpendicular to the longest diagonal, from the all-zeros corner to the all-ones corner. This is the cross-section corresponding to the plane $x_1 + \cdots + x_n = n/2$.

Alternatively, then, you want the probability density function of the sum $S_n$, of $n$ independent uniform(0,1) random variables, evaluated at $n/2$.

By the central limit theorem, the $S_n$ have asymptotic normal distribution with mean $n/2$ and variance $n/12$. So the answer should behave asymptotically like $1/\sqrt{2\pi n/12}$, i. e. like $\sqrt{\pi/6n}$.

It appears that this distribution is called the Irwin-Hall distribution.