Gauss theorem in integral form gives rather an intuitive answer. Flux of electric field through a surface $D$ $\Phi=\int_D \vec E\cdot \mathrm{d} \vec A$ is proportional to the total charge $Q$, contained inside the surface. Now, compare three cases:
1) Point source. Draw spherical shells centered on the point source. Due to spherical symmetry $\vec E$ is normal to the shells and depends only on the radius of a shell. Hence $Q\sim \Phi=\int_D \vec E\cdot \mathrm{d} \vec A= E \int_D \mathrm{d} A=E A$. As the area of spheres goes as $A\sim r^2$, one has $E\sim A^{-1}\sim r^{-2}$.
2) Line source. Draw cylinders aligned with the source. Here the symmetry of the problem tells us that $\vec E$ is normal to the cylinders, and again depends only on the radius of a cylinder. Derivation is the same as before, except for that the area of a cylinder $A\sim r$ instead of $r^2$. Hence $E\sim r^{-1}$.
3) In the case of a plane, draw prisms with the base parallel to the plane and having the same area, so that each prism crosses the plane. Due to symmetry reasons the field is normal to the plane, and hence to the bases of the prysms. As the area of the base is the same for all the prisms, the charge inside the prisms is also the same. On the other hand $A$ doesn't depend on the height of the prisms at all. Hence $E\sim r^0$.
The answer to your question then, according to Gauss theorem, is that the electric field is inversely proportional to the area of the surfaces, over which the field is constant.
Your friend is not entirely right, as you have noticed, though he has a good point: wherever from you look at an infinite plane, it always looks infinite, and hence has the electric field normal to it at any point in space.
