Magnetostatic energy density -- derivation without introducing inductance?

It's a reasonable question, and the answer is: one can't prove it, without introducing induction.
Consider a conducting loop with no current. Then someone starts creating a current in it, using, for example, a battery. The question is: why should we perform work to create this current, if we know that magnetic force $$ \mathbf {F} = \frac{q}{c}[\mathbf{v},\mathbf{B}]$$ is always perpendicular to the charge's displacement (thus it doesn't perform work at all)? The answer for this is that when current appears, magnetic field changes, thus electric field is created. Electric field acts upon charges with the force $$ \mathbf {F} = q\mathbf{E}$$ and because of that we must take an effort to create the current.
Moreover, it holds true in general case. We must $\textit {always}$ work against electric field in order to create magnetic field. In some sense, it is meaning of magnetic energy.
The best way to introduce magnetic and electric energy is Poynting theorem. It states: $$\mathbf{E}\cdot\mathbf{j}+\frac{\partial u}{\partial t}=-\triangledown\mathbf{S}$$ where $u=\frac{E^2}{8\pi}+\frac{B^2}{8\pi}$, $\mathbf{j}$ is current density and $\mathbf{S}$ is Poynting vector. I strongly recommend you to read a corresponding Feynman's lecture, you will defenetly get the sense of these things.

$\textbf{PS:}$ I used CGS system here, I hope it didn't bug you very much.

You can certainly do the derivation without resorting to the inductance. When it comes about to consider the contributions of induced electromotive forces $V_i$ you can consider that $$ V_i=-\frac{d\Phi}{dt} $$ where $\Phi$ is the flux of the magnetic field $B$ through the surface of the circuit you're considering. You can find a derivation according to this line of reasoning in the text from Stratton.