Let $n\in\mathbb{N}$ and $a\in\mathbb{R}^n$. The question is to find the value $$A_n := \int_{S_{n-1}}\exp(i a\cdot\xi)\,\mathrm{d}S(\xi),$$ where $S_{n-1}$ denotes the $(n-1)$-dimensional sphere in $\mathbb{R}^n$, and $\mathrm{d}S$ stands for surface integration, and $a\cdot\xi$ is the scalar product of $a$ and $\xi$.
For $n=2$, the value is $A_2 = 2\pi J_0(\lvert a\rvert)$, with $J_0$ the Bessel function of zeroth order.
I was able to compute this. The general answer is $$A_n = 2^{n/2} \pi^{n/2} \frac{J_{n/2-1}(\lvert a\rvert)}{{\lvert a\rvert}^{n/2-1}}.$$ This follows from using the orthogonal invariance of the surface measure on the sphere (thus it suffices to consider $a = \lvert a\rvert(1,0,\ldots,0)$) and then expanding the exponential function. I'll elaborate if there is interest.