As mentioned in NotAstronaut's answer, objects smaller than 25 meters will typically burn up in the atmosphere. One can very easily see why this should be the case using Newton's impact depth formula. This is based on approximating the problem by assuming that the matter in the path of the object is being pushed at the same velocity of the object, so as soon as the object has swiped out path containing the same mass as its own mass, it will have lost all of its initial momentum. All its kinetic energy will then have dissipated there, so if this happens in the atmosphere it will have burned up before reaching the ground.
This is, of course, a gross oversimplification, but it will yield correct order of magnitude estimates. We can then calculate the critical diameter as follows. The mass of the atmosphere per unit area equals the atmospheric pressure at sea level divided by the gravitational acceleration, so this is about $10^4\text{ kg/m}^2$. If an asteroid of diameter $D$ and density $\rho$ is to penetrate the atmosphere, its mass of $1/6\pi \rho D^3$ should be larger than the mass of the atmosphere it will encounter on its way to the ground, which is $5/2 \pi 10^3 D^2\text{ kg/m}^2$. Therefore:
$$ D > \frac{1.5\times 10^4}{\rho} \text{ kg/m}^2$$
If we take the density $\rho$ to be that of a typical rock of $3\times 10^3 \text{ kg}/\text{m}^3$, then we see that $D>5\text{ m}$, which is reasonably close order of magnitude estimate to the correct answer.