There should be a relationship like this, because the heat function proof of Hirzebruch-Riemmann-Roch uses a much more refined equality between hilbert functions and Laplacian spectra. Let me see if I can put this together.
Let $M$ be a $d$-dimensional smooth projective variety. Let $L^k$ be the line bundle on $M$ obtained by restricting $\mathcal{O}(k)$ from projective space. Equip $M$ with the Fubini-Study metric. We have the complex of vector spaces:
$$0 \to C^{\infty}(L^k) \overset{\bar{\partial}}{\longrightarrow} C^{\infty}(L^k \otimes \Omega^{0,1}) \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} C^{\infty}(L^k \otimes \Omega^{0,d}) \to 0. \quad (\dagger)$$
Here $\Omega^{0,q}$ is the vector bundle of $(0,q)$-differential forms and $C^{\infty}(\mathrm{vector \ bundle})$ means $C^{\infty}$ sections of that vector bundle. We can use the metric on $L^k$ to define a Hodge star operator on sections of $L^k \otimes \Omega^{0,q}$. Let $\bar{\partial}^{\ast}$ be the adjoint to $\bar{\partial}$ and let $\Delta_{L^k} = \bar{\partial} \bar{\partial}^{\ast} + \bar{\partial}^{\ast} \bar{\partial}$. This is not the ordinary Laplacian -- it acts on sections of $L^k$ tensored with differential forms rather than acting on differential forms. All of this is pretty standard : See Wells Differential Geometry on Complex Manifolds or Voisin's Complex Algebraic Geometry and Hodge Theory for background on this kind of construction.
Since $\bar{\partial}$ commutes with $\Delta_{L^k}$, the sequence $(\dagger)$ splits up into finite dimensional sequences for every eigenspace of $\Delta_{L^k}$. (There is some deep analysis with Sobolev spaces necessary to make this precise.) Let $(\dagger)_{\lambda}$ be the corresponding sequence of $\lambda$ eigensequences. Since $\Delta_{L^k}$ is positive semidefinite, you only get terms with $\lambda \geq 0$.
For $\lambda >0$, we have the homotopy $\mathrm{Id} = (1/\lambda) (\bar{\partial} \bar{\partial}^{\ast} + \bar{\partial}^{\ast} \bar{\partial})$, so $(\dagger)_{\lambda}$ is exact for $\lambda>0$. On $(\dagger)_0$, we have $\bar{\partial}=0$ so the sequence is trivial. For $k$ large, it turns out that only the first term has a nontrivial $0$-eigenspace. That eigenspace is the kernel of $\bar{\partial}$ acting on $C^{\infty}(L^k)$, which is to say, the holomorphic sections of $L^k$. So its dimension is the Hilbert function. More generally, for all $k$, the alternating sum of the dimensions of the $0$-eigenspaces is the Hilbert polynomial.
Gather up the eigenvalues into a generating function: $$\theta(L^k, q, t) = \sum e^{-\lambda^2 t} \dim{\LARGE (}\lambda\mathrm{-eigenspace \ of\ } C^{\infty}(L^k \otimes \Omega^{0,q}) {\LARGE )}.$$ Then the above argument shows that $$h(k) = \sum_q (-1)^q \theta(L^k, q, t). \quad (\S)$$ Note that $(\S)$ holds for all $t$.
Now, in the height function proof, one engages in a detailed analysis of the asymptopics of $\theta(L^k, q, t)$ as $t \to 0$. When $t \to 0$, all of the $e^{-\lambda^2 t}$ terms go to $1$, so $\theta$ blows up, and the rate at which it blows up depends on the growth rate of the eigenvalues of $\Delta_{L^k}$. One gets an asymptopic formula that looks like $\Theta(t) = a_{d} t^{-d/2}+a_{d-1} t^{-(d-1)/2} + \cdots + a_0 + O(t^{1/2})$. Plugging into $(\S)$ and comparing constant terms, one gets a formula for $h(k)$.
This computation is really complicated, involving integrals of products of chern classes of $M$ with the powers of the Fubini-Study form. You are only interested in leading behavior for $k$ large, and you want to make the transition to thinking about $\Delta$ rather than $\Delta_{L^k}$.
I haven't seen anyone work this out, but I do remember there is a lemma which relates $\Delta$ and $\Delta_{L^k}$ for $k$ large and that Griffiths and Harris use it in their proof of Serre vanishing. I covered this in my Hodge theory course, so I am a little embarrassed not to remember the details! It should be in the notes for April 12. Anyway, possibly you can figure out how to finish the proof from here.