no, it is not positive everywhere; notice that the integral only depends on $|x|$, so we may orient the vector $x$ along the $x_1$ axis; going to hyperspherical coordinates we have
$$\Phi_n(x)=\frac{2\pi^{(n-1)/2}}{\Gamma[\tfrac{1}{2}(n-1)]}\int_0^\infty dr \int_0^\pi d\phi\; e^{-r^n}\cos\bigl(|x|r\cos\phi\bigr)r^{n-1}(\sin\phi)^{n-2}$$
I checked that it becomes negative for $n=3$:
$$\Phi_3(x)=\frac{4\pi}{|x|}\int_0^\infty e^{-r^3}r\sin(|x|r)\,dr$$
http://ilorentz.org/beenakker/MO/phi_3.png
no, it is not positive everywhere; notice that the integral only depends on $|x|$, so we may orient the vector $x$ along the $x_1$ axis; going to hyperspherical coordinates we have
$$\Phi_n(x)=\frac{2\pi^{(n-1)/2}}{\Gamma[\tfrac{1}{2}(n-1)]}\int_0^\infty dr \int_0^\pi d\phi\; e^{-r^n}\cos\bigl(|x|r\cos\phi\bigr)r^{n-1}(\sin\phi)^{n-2}$$
I checked that it becomes negative for $n=3$:
$$\Phi_3(x)=\frac{4\pi}{|x|}\int_0^\infty e^{-r^3}r\sin(|x|r)\,dr$$ http://ilorentz.org/beenakker/MO/phi_3.png
no, it is not positive everywhere; notice that the integral only depends on $|x|$, so we may orient the vector $x$ along the $x_1$ axis; going to hyperspherical coordinates we have
$$\Phi_n(x)=\frac{2\pi^{(n-1)/2}}{\Gamma[\tfrac{1}{2}(n-1)]}\int_0^\infty dr \int_0^\pi d\phi\; e^{-r^n}\cos\bigl(|x|r\cos\phi\bigr)r^{n-1}(\sin\phi)^{n-2}$$
I checked that it becomes negative for $n=3$:
$$\Phi_3(x)=\frac{4\pi}{|x|}\int_0^\infty e^{-r^3}r\sin(|x|r)\,dr$$
