When can a molecule be considered freely rotating at room temperature?
I'm not a transition-state physical chemist, but I think a good approach to this problem is transition-state theory, specifically the Eyring equation:

$$k = \kappa \frac{k_b T}{h} e^{\frac{-\Delta G^{\ddagger}}{RT}}$$

This equation tries to predict the rate constant $k$ from an assumed pseudo-equilibrium between the transition-state and the starting material, which are assumed to differ in energy by $\Delta G^{\ddagger}$.  $k_b$ is Boltzmann's constant, $h$ is Planck's constant, $T$ is temperature, and $\kappa$ is the "fudge factor" or transmission coefficient, the fraction of transition states that react productively instead of converting back to starting material.

In the conformational reactions you are talking about, I think a good starting place would be to assume the eclipsed conformations are the transition state, although I don't know if this has been shown to be 100% true experimentally or computationally.   As a first approximation, let's assume there isn't a huge entropy term in $\Delta G^{\ddagger}$, so that it is well-approximated by the energy level differences you found.  If we also assume that $\kappa$ is 1.0, then we can just multiply the $e^{\frac{-\Delta E}{RT}}$ you've already calculated by $\frac{k_b T}{h} = 6.2\times10^{12}\mathrm{~s^{-1}}$ at 298 K to get (very) approximate rate constants.


Ethane: $6.55 \times 10^{-3} \rightarrow 4.1 \times 10^{10}\mathrm{~s^{-1}}$
Butane: $4.79 \times 10^{-4} \rightarrow 3.1 \times 10^{9}\mathrm{~s^{-1}}$
$\ce{H2O2}$: $4.29 \times 10^{-5} \rightarrow 2.0 \times 10^{8}\mathrm{~s^{-1}}$
Ethene: $2.09 \times 10^{-44} \rightarrow 1.3 \times 10^{-31}\mathrm{~s^{-1}}$


That shows that rotational equilibrium of everything except ethene should be very rapid, but that the half-life for ethene equilibration much greater than the age of the universe ($4 \times 10^{17} \mathrm{~s}$).

Suppose we want to find an energy difference that should lead to a rate constant that is neither very fast nor very slow, say we'd like $k\approx\mathrm{1~hr^{-1}}$.  Inverting our very very crude approximation gives:

$$\Delta E = - RT \ln{\left(\frac{k h}{\kappa k_b T}\right)}$$

Plugging in $k=\mathrm{1~hr^{-1}=2.7\times10^{-4} s^{-1}}$ gives (if I did my math right) 93 kJ per mol.  Some well-known atropoisomers do have estimated energy differences in that range.