Superfluid helium 4 is the winner here, and by a vast margin. There's a body of literature from the 90s that attempts to estimate the fraction of helium atoms that are Bose condensed in superfluid helium 4. To quote this paper, "These works... estimate the condensate fraction to be ≈7% at saturated vapor pressure (SVP)." At a mass density of 125 g/L, this would give about $10^{24}$ condensed atoms per liter!
Note that while helium liquifies below 4K, you need to go below ~2K to get superfluid helium. The transition from liquid to superfluid is continuous, and so the amount of superfluid helium right at the transition temperature is basically zero. However below about 1K, virtually all helium atoms are in the superfluid state.
That an Avogadro's number or more of atoms can occupy the zero-momentum state of an arbitrarily large box is a pretty remarkable demonstration of Bose-Einstein condensation.
Honorable mentions
Regarding more conventional BECs, the largest I've come across is only slightly larger than the 1 billion you quote in your question; $1.1 \pm 0.6 \times 10^9$ atoms in an earlier paper by the same authors. Most research these days works with only millions of atoms (and rarely hydrogen), so I'm not aware of any efforts to push above billions.
Superconductors have a lot in common with Bose Einstein condensates, since in this effect pairs of fermions also condense into a macroscopic 'supercurrent.' No two fermions can occupy the same state, but if you'll accept the similarities, then there are some theoretical proposals that entire stars can form a "BEC". See e.g. Bose-Einstein Condensate general relativistic stars, though I can't speak to the correctness of this paper.
Edit: Wait, really???
Some of the comments are understandably skeptical of the claim in this answer. As OP points out (essentially), a helium atom confined to a 1 mm $\times$ 1 mm $\times$ 1 mm box would have a ground state energy of $E=3\hbar^2 \pi^2/2mL^2 = 2.5\times 10^{-35}$ Joules, which corresponds to a temperature of
$$T_\text{ground state} = E/k_b = \frac{3\hbar^2\pi^2}{2m L^2 k_B} = 1.8 \times 10^{-12}~K$$
How then can it be that liquid helium atoms would occupy the ground state of an arbitrarily large box at 2 $K$ or more?
But the magic of Bose-Einstein condensation is that a dense collection of particles will start to pile up in the ground state of a potential well at much higher temperatures than would an isolated particle of the same kind. The critical temperature for a Bose-Einstein condensate in 3D is
$$T_c = \frac{\hbar^2 \rho_n^{2/3}}{m k_b} = \frac{\hbar^2 N^{2/3}}{mL^2 k_B}$$
where $\rho_n$ is the particle density and $N$ is the total number of particles. Notably these two expressions differ by the factor of $N^{2/3}$ in the numerator, which makes $T_c$ much much larger than $T_\text{ground state}$. If we plug in the number density for liquid helium (125 $kg/m^3 \rightarrow 1.8\times 10^{28}~ \text{particles}/m^3$), then we get a critical temperature of 0.8 $K$. This agrees quite well with the actual critical temperature of 2 $K$, despite the fact that we incorrectly treated the helium atoms as non-interacting particles.
