Why does field strength follow the inverse square law but potential does not?

In the A-Level, you are already working with single variable calculus; if you are working in multivariable calculus and know what a gradient is, this will make more sense, so please comment if you do.

The correct equation to consult here would be the relationship between a force and a potential: energy:

$$F = -frac{dW}{dx}$$

Intuition: A field exerts a force on a particle on the direction of maximum decrease of energy (because, empirically, we found that the universe wants everything to have the least potential energy it possibly can, at least in classical physics). In the single variable case this “direction” can be either $+x$ or $-x$, however, in the case where we have more than 1 dimension, this is much more illuminating: suppose you place a ball on a hill.Where will it go? You will find it accelerates (I.e feels a force) in the direction of steepest descent. This thinking applies analogously to electrostatics.

For a given particle, the relationship between electric potential energy and the electric potential is $$ W = q V(x) $$ where $q$ is the charge of the particle, $W$ is the potential energy and $V$ is the potential. This one is less fundamental: The potential is a property of the field, but two particles in the same $x$ would have the same potential but different energies, depending on their charge.

For completeness, the relationship between a force exerted by the electric field on the particle, and a electric field, is $$ F= qE(x)$$ Because, in similar fashion to before, two particles with different charges in the same point in space should feel a force proportional to their charge.

With these three equations in mind, we should see that the relationship between an electric field and a electric potential is a derivative:

$$ E = -frac{dV}{dx} $$ and thus an inverse square force/electric field should correspond to an inverse potential.

One can look at it as follows: The field has to fall off as $1/r^2$ because the flux through a spherical surface is independent of distance, and of course, the force is proportional to the field:

$$ vec F = qvec E $$

The energy required to get a charge from $infty$ to $r$ is computed from the work done along the path:

$$ W(r) = int_{infty}^r{F(r')dr'}= int_{infty}^r{frac{q}{r'^2}dr'}=-frac q r$$

which is also called the potential.