Significance of electrical fields of infinite objects
I don't like the "you can't get away" explanation. There is a simple explanation with field lines:
In all three cases, the field lines are straight lines from the point charge to infinity. You can easily calculate the density of the field lines for each object. For a point charge, the "number" of field lines through any sphere around the point charge is the same (as all field lines are straight, they have to pass any sphere centered at the charge exactly once). The density is $n/A$, where $n$ is the number of lines and $A$ is the surface area of the sphere. Therefore, since the number of lines is constant and the surface of the sphere is $r^2$, the density goes as $1/r^2$.
For a line, the density only thins out around circles of the line. In other words, the number of field lines through each cylinder around the line is the same. Since the area of the cylinder is proportional to $r$, the density of field lines grows as $1/r$.
Finally, for the plane, the number of lines through any plane parallel to the plane is the same. Since the surface doesn't change, the density is constant (this is also true, because all field lines are parallel).
The only question that remains is why the field strength is proportional to the density of the field lines. Why does that work? I guess it's somehow intuitive, but for a technical answer let me refer to the excellent answer by Emilio Pisanty here: Why does the density of electric field lines make sense, if there is a field line through every point?
I believe the answer to your question lies in the Gauss theorem itself $$ \oint \textbf{E}d\textbf{S} \sim Q $$ and the symmetry of the system, which defines the shape of equipotential surfaces.
In case of a point charge there is a rotational symmetry about any axis going through the charge, so the equipotential surfaces are spheres whose area is proportional to $r^2$. Thus the field is $\sim r^{-2}$ ($r$ is distance from the charge)
In case of a line, rotational symmetry exists only around this line, so equipotential surfaces are cylinders, whose area (per length along the line) is $\sim r$ (here $r$ is the shortest distance to the line). Thus the field is $\sim r^{-1}$
For a plane, equipotential surfaces are planes, parallel to the given plane, their area (again, per unit area of the given plane) is independent of the distance from the given plane, as is the field.
The case of sphere is not like the ones above! Here you are inside the sphere, so the charge $Q$ on the rhs of gauss law also changes with the distance from the center, proportionally to the volume $\sim r^3$. Equipotential surfaces are spheres again, whose area is $\sim r^2$, so the field increases linearly with $r$. If you were outside the sphere the field would decrease $\sim r^{-2}$ as in case 1.