Though it does not have the direct relation to the formulated task, it would be interesting to get a closed form of $\displaystyle I(\xi,n)=\int_{\mathbb{R}^n} \frac{{\lvert e^{i\langle\vec\xi,\vec y\rangle}+e^{-i\langle\vec\xi, \vec y\rangle}-2\rvert}^2}{\lVert \vec y \rVert^{n+2}}dy$. Denoting $\displaystyle\xi=|\vec\xi|$ and $\displaystyle r=|\vec y|$
$$I(\xi,n)=2^4\int_{S^{n-1}}d\Omega \int_0^\infty\frac{\sin^4\frac{\xi \,r\cos\phi_1}{2}}{r^3}dr=4\xi^2\int_{S^{n-1}}\cos^2\phi_1d\Omega\int_0^\infty\frac{\sin^4x}{x^3}dx\tag{1}$$
For n-sphere $d\Omega$ is defined as $\displaystyle d\Omega=\sin^{n-2}\phi_1\sin^{n-3}\phi_2...\sin\phi_{n-2}\,d\phi_1...d\phi_{n-2}$ ( here)
The interval of integration over $d\phi_1$ is $[0;\pi]$, therefore
$$\int_0^\pi\cos^2\phi_1\sin^{n-2}\phi_1d\phi_1=\int_0^\pi\sin^{n-2}\phi_1d\phi_1-\int_0^\pi\sin^n\phi_1d\phi_1$$
Integrating the second term by part,
$$\int_0^\pi\sin^n\phi_1d\phi_1=(n-1)\int_0^\pi\sin^{n-2}\phi_1\cos^2\phi_1d\phi_1=(n-1)\int_0^\pi\sin^{n-2}\phi_1(1-\sin^2\phi_1)d\phi_1$$
$$\int_0^\pi\sin^n\phi_1d\phi_1=\frac{n-1}{n}\int_0^\pi\sin^{n-2}\phi_1d\phi_1$$
$$\int_0^\pi\cos^2\phi_1\sin^{n-2}\phi_1d\phi_1=\frac{1}{n}\int_0^\pi\sin^{n-2}\phi_1d\phi_1$$
and, therefore,
$$\int\cos^2\phi_1d\Omega=\frac{1}{n}\int d\Omega\tag{2}$$
To evaluate the last integral, we consider
$$\int_{-\infty}^\infty...\int_{-\infty}^\infty dx_1...dx_n e^{-x_1^2-...-x_n^2}=\pi^\frac{n}{2}=\int d\Omega\int_0^\infty e^{-r^2}r^{n-1}dr$$
$$=\frac{1}{2}\int d\Omega\int_0^\infty e^{-t}t^{n/2-1}dt=\frac{1}{2}\Gamma\Big(\frac{n}{2}\Big)\int d\Omega\,\,\Rightarrow\,\,\int d\Omega=\frac{2\pi^\frac{n}{2}}{\Gamma\big(\frac{n}{2}\big)}$$
and
$$\int_{S^{n-1}}\cos^2\phi_1d\Omega=\frac{1}{n}\int_{S^{n-1}} d\Omega=\frac{2\pi^\frac{n}{2}}{n\,\Gamma\big(\frac{n}{2}\big)}\tag{3}$$
Now we only have to evaluate $\displaystyle J=\int_0^\infty\frac{\sin^4x}{x^3}dx$. Integrating a couple of times by part,
$$J=\int_0^\infty\frac{\sin2x\sin^2x}{x^2}dx=\int_0^\infty\big(2\cos2x\sin^2x+\sin^22x\big)\frac{dx}{x}$$
$$=\int_0^\infty\frac{\cos2x-\cos4x}{x}dx=\Re\,\int_0^\infty\frac{e^{2ix}-e^{4ix}}{x}dx$$
Integrating along a closed contour the complex plane (formally, it means the substitution $x=it$),
$$J=\int_0^\infty\frac{e^{-2t}-e^{-4t}}{t}dt=\ln2\tag{4}$$
(in the last step we used the Frullani theorem)
Putting (3) and (4) into (1)
$$\boxed{\,\,I(\xi,n)=\frac{8\ln2}{n}\frac{\pi^\frac{n}{2}}{\Gamma\big(\frac{n}{2}\big)}\xi^2\,\,}$$
It is easy to check independantly that, for example, at $n=3$ it gives the correct result:
$$ I(\xi, n=3)=\frac{16\pi\ln2}{3}\xi^2$$
