Joel is right that it is partly just a convention to scale Gaussian curvature so that the curvature of a unit sphere is $1$. However, there are three natural motivations for this scale besides matching 1 to 1 in the case of a sphere.

First, Gauss defined his curvature as the product of the extrinsic curvatures of a surface in $\mathbb{R}^3$. So there is a coefficient of 1 in this natural formula.

Second, the unit sphere has the property that the deviation from Euclid's parallel postulate has a factor of 1. In other words, the area $A$ of a triangle with angles $\alpha, \beta, \gamma$ is $\alpha + \beta + \gamma - \pi$. In general, if you have a very small triangle with area $A$ at a point of local curvature $K$, its angle deviation is $KA$ to first order. This factor of 1 leads to a factor of $2\pi$ in the Gauss-Bonnet theorem, that the integral of Gaussian curvature is $2\pi \chi$.

Third, Gaussian curvature is the ratio of the Ricci curvature tensor to the metric, and it is also half of the scalar curvature.

In comparing these formulas, the most reasonable scales for Gaussian curvature are the standard choice, the standard choice times 2 to match scalar curvature, and the standard choice divided by $2\pi$ to match the Gauss-Bonnet theorem. The volume and area formulas are some justification for a 1/3 or a 1/12 or similar, but these are taken to be less fundamental scales.

(One irony of the discussion is that $\pi$ itself is half of the most important value in trigonometry.)

Also the volume and surface area ratios are given in Wikipedia in $n$ dimensions. It is also worth looking at the generalized Gauss-Bonnet theorem in $2n$ dimensions.