No, I don't believe there is. Or, describing the scope of my answer, there is no maximum "metallicity" (for any normal mixture of metals) that could prevent a collapsing protostar becoming hot enough in its core to initiate nuclear fusion.
(If your question is about the Jeans mass and metallicity, then you could clarify).
What determines whether fusion will ever commence is whether the contraction of the protostar is halted by electron degeneracy pressure before reaching a temperature sufficient for nuclear ignition.
For a solar composition protostar, the critical mass is about $0.08M_{\odot}$. Below this, the core does not attain a temperature of $\sim 5\times 10^{6}$ K that are required for nuclear fusion.
The calculation of this minimum mass depends on $\mu_e$, the number of mass units per electron in the core (which governs electron degeneracy pressure), and on $\mu$, the number of mass units per particle in the core (which governs perfect gas pressure). However, these dependencies are not extreme. In the core of the protosun, $\mu_e \sim 1.2$ and $\mu \sim 0.6$. If we made a metal rich star that had very little hydrogen by number and the rest say oxygen (a.k.a. a star made of water), then $\mu_e \sim 1.8$ and $\mu \sim 1.6$. The minimum mass for hydrogen fusion is given approximately by
$$ M_{\rm min} \simeq 0.08 \left( \frac{\mu}{0.5} \right)^{-3/2} \left(\frac{\mu_e}{1.2}\right)^{-1/2}$$
(e.g., see here).
These different parameters would be enough to change the minimum mass (downwards actually) for hydrogen fusion to around $0.012 M_{\odot}$.
We could of course hypothesise a star that was wholly made of metals. A convenient estimate of the minimum mass for carbon fusion is already supplied by stellar evolution models. A $>8M_{\odot}$ star with a carbon core will initiate carbon fusion before it becomes degenerate. The mass is much higher than for H fusion because of the increased coulomb barrier between carbon nuclei. Of course the star also has a hydrogen/helium envelope, but if you replaced this with carbon, then the result will be little changed. Thus you could have a population of lower mass objects that do not become stable "stars". Those with masses of $1.4 < M/M_{\odot}
< 8$ would presumably end up detonating as some kind of type Ia supernovae, because they will achieve a density/temperature combination where C can fuse, but in highly degenerate conditions. Lower than that and it becomes a stable white dwarf.
Of course your metal rich "star" could just be a ball of iron, in which case nuclear fusion isn't going to happen and if it is more than $\sim 1.2M_{\odot}$ it will collapse directly to a neutron star or black hole, possibly via some sort of supernova. Lower than that and it becomes a stable iron white dwarf.
Your suspicion is correct, the Bonnor-Ebert mass is indeed describing the same instability condition as the Jeans instability. This is the gravitational instability, which is tied to the competition between self-gravity and internal pressure. The only reason they are given different names is that they represent two different ways to arrive at this instability condition, which end up giving compatible results.
To derive the Bonnor-Ebert mass, one considers equilibrium solutions to a spherically symmetric configuration of self-gravitating gas in hydrostatic equilibrium. These are solutions to the Lane-Emden equation. One can then consider the stability of normal modes in these solutions to perturbations. For an isothermal equation of state, the fundamental ("breathing") mode is unstable whenever the mass of the sphere exceeds
$$ M_{BE} = 1.18 \frac {c_s^3}{\rho_0^{1/2} G^{3/2}}$$
where $c_s$ is the isothermal sound speed and $\rho_0$ is the central density of the sphere. In other words, spherical configurations of gas with isothermal equations of state that are initially in equilibrium are unstable to collapse when perturbed if they are more massive than $M_{BE}$. This analysis can be extended to gas with non-isothermal equations of state.
To derive the instability from the Jeans point of view, one can instead consider traveling-wave perturbations (sound waves) traveling through a self-gravitating, homogeneous medium of density $\rho_0$ and isothermal sound speed $c_s$. Through linear analysis of small amplitude waves one can derive the dispersion relation for these waves and conclude that when the wavelength exceeds a critical length,
$$\lambda_J = \sqrt{\frac{\pi c_s^2}{G \rho_0}}\, ,$$
the amplitude grows exponentially in time. So structures that have lengths larger than this are susceptible to collapse when subjected to perturbations. Again, this was all done in the isothermal case for simplicity, but can be generalized to other equations of state.
To see how these two analyses relate to each other, consider the mass enclosed within a sphere of diameter $\lambda_J$ and uniform density $\rho_0$. You'll see that it is the same as $M_{BE}$ to within a factor of 2 or so, which shouldn't be too worrisome given the differences in the initial configurations (an equilibrium sphere for $M_{BE}$, and a uniform medium for the Jeans analysis).
For a detailed discussion, see e.g. chapter 9 of this textbook. The primary application there is star formation, but the physics is general and may be applied to other situations.
Best Answer
Short answer: gravitational potential energy is converted into heat.
Let's look at the Sun as an example. Its mass is $M_\odot = 2.0\times10^{30}\ \mathrm{kg}$ and its radius is $R_\odot = 7.0\times10^8\ \mathrm{m}$. If its density were uniform, its gravitational binding energy would be $$ U_{\odot,\,\text{uniform}} = -\frac{3GM_\odot^2}{5R_\odot} = -2.3\times10^{41}\ \mathrm{J}. $$ In fact the Sun's mass is centrally concentrated, so $U_{\odot,\,\text{actual}} < U_{\odot,\,\text{uniform}}$.
Where did the Sun come from? Something like a giant molecular cloud with a density of $2\times10^{-15}\ \mathrm{kg/m^3}$. The mass of the Sun would thus have been extended over something like a sphere of radius $6\times10^{14}\ \mathrm{m}$, for a gravitational binding energy of $$ U_\text{cloud} = -3\times10^{35}\ \mathrm{J}, $$ which is negligible in comparison with $U_\odot$.
All of the $2.3\times10^{41}\ \mathrm{J}$ had to go somewhere, and the only place to dump energy is into heat. The gas particles gain velocity as they fall into the potential well, but they don't lose that velocity because they never climb back out of the well.
Not worrying about whether the heating is isobaric or isochoric or somewhere in between, the heat capacity of monatomic gas is about twice the ideal gas constant, or $8.3\times10^3\ \mathrm{J\,K^{-1}\,kg^{-1}}$. At this amount, in order to heat all of $M_\odot$ by the average temperature of the Sun (say $10^7\ \mathrm{K}$, somewhere between the core and surface temperatures), you would need about $1.7\times10^{41}\ \mathrm{J}$ of energy. There is enough energy released by gravitational collapse to heat the Sun to its current temperature. You can do a more detailed analysis taking into account how much cooling occurs during the collapse, but the steep temperature dependence of the Stephan-Boltzmann law makes it difficult to lose heat to space until the object is already hot. I'm also neglecting a factor of $2$ that comes from splitting the energy between heating the gas and compressing it.
Once the material is this hot, it simply glows like any blackbody emitter. The energy lost to space is replenished by nuclear fusion in the core. In fact, fusion acts as a regulator: too much of it and the star expands and cools, slowing down fusion; too little and the star collapses further, heating up more and increasing the fusion rate.
In summary, gravitational collapse provides the initial energy to heat a star. As it uses up this energy source, it begins to tap into fusion. Ultimately it reaches an equilibrium where the energy produced by fusion is balanced by the energy radiated into space.