Let $F(x_1, \dots, x_n)= (x_1, \dots, x_n)$. From the Divergence Theorem we have
$$\int_{B_1(0)} \nabla \cdot \mathbf{F} \space dV = \int_{\partial B_1(0)} \mathbf{F} \cdot \mathbf{\nu} \space dS, $$
where $\mathbf{\nu}$ is the normal unit vector to $\partial B_1(0)$.
Hence we have
$$\int_{B_1(0)} \nabla \cdot \mathbf{F} \space dV = n\int_{B_1(0)} \space dV = n \cdot \mathcal{L}^n (B_1(0)) = \int_{\partial B_1(0)} dS = \mathcal{L}^{n-1} (\partial B_1(0)),$$
where $\mathcal{L}^n$ denotes the $n$-dimensional Lebesgue measure. So we proved that the measure of the unit ball in $\mathbb{R}^n$ is equal to the the measure of its surface divided by the dimension $n$.