This is the final revision and probably my final post for a long time (deadlines). I wanted to write a more definitive answer, about how to approach the Hodge theorem, since I've thinking about it for a while. This a bit long. So the short summary is that there is no really easy path, but each is beautiful in it's own way.
Method 1 (orthogonal projections): This is the standard proof, although there are many variations which can be found in Griffiths-Harris, Warner, Wells,... Here's a very rough idea. The basic problem is to show that the space $closed(X)$ of closed $C^\infty$ forms on a compact Riemannian manifold $X$ is a direct sum of the space of exact forms $exact(X)$ and the space of harmonic forms $harm(X)$. A closed form is harmonic if and only if it is orthogonal to closed form (easy), so one might try to first prove that
$$closed(X)= exact(X)\oplus exact(X)^\perp$$
and then identify the latter with $harm(X)$. Since these are infinite dimensional, the decomposition isn't automatic. However, one can make it work by using $L^2$ forms and applying Hilbert space methods to obtain:
$$\overline{closed(X)}= \overline{exact(X)}\oplus \overline{exact(X)}^\perp$$
But at end one wants to come back to $C^\infty$ forms, and here is where the magic of elliptic operators comes in. The basic result which makes this work is the regularity theorem: a weak solution of elliptic equation, e.g Laplace's equation, is in fact a true $C^\infty$ solution. The space on the right of the second decomposition is therefore $harm(X)$, and this is all one needs.
As you said, the full details entail quite a bit of work, but one can look at standard books on Riemann surfaces (Farkas-Kra, Forster, Narasimhan...) for some instructive special cases.
Method 2 (Heat Equation): The heuristic is that if you think of the closed form as an initial temperature, governed by the heat equation, it should evolve toward a harmonic steady state. The nice thing is that this can solved explicity on Euclidean space, and this gives a good (short time) approximation for the general case. If I may make a shameless plug, I wrote up an outine in chapter 8 of my notes
http://www.math.purdue.edu/~dvb/preprints/book.pdf
Method 3 (Deligne-Illusie): This is really an amplification of Kevin Lin's answer. One important consequence of the Hodge theory is the degeneration of the Hodge to De Rham spectral sequence $$H^j(X,\Omega_X^i) \Rightarrow H^{i+j}(X,\Omega_X^\bullet) = H^{i+j}(X,\mathbb{C})$$ when $X$ is smooth and projective along with Kodaira vanishing. The first algebraic proof of this was due to Falting (on the way to Hodge-Tate). Deligne and Illusie gave a comparatively elementary proof. Although as Ravi Vakil commented, it is not the best way to first learn this stuff. Nor does it give the full Hodge decomposition. However, for people who want to go this route, an introduction aimed at students can be found in the book by Hélène Esnault and the late Eckart Viehweg.
Addendum (added June 24): I wanted to briefly address the question of how much of the Hodge decomposition can be understood algebraically. One can define algebraic de Rham cohomology with its Hodge filtration coming from the spectral sequence above. What is missing is a purely algebraic description of the "Betti lattice" $image [H^i(X,\mathbb{Z})\to H^i(X,\mathbb{C})]$ and the fact that the conjugate filtration $\overline{F}$ and $F$ are opposed in Deligne's sense. The first issue already seems serious. Even if $X$ is defined over $\mathbb{Q}$, a basis for the lattice typically involves transcendentals. This is already clear in the simplest example $H^1(\mathbb{A}^1-\{0\})$, the lattice is spanned by $[\frac{dz}{2\pi iz}]$.