What was the age of the universe when the average density was one atmosphere?
The critical density of the universe is 
$$\rho_c=\frac{3H_0^2}{8\pi G},$$
with $H_0=67.8 \frac{\mbox{km}/\mbox{s}}{\mbox{Mpc}}$ the Hubble constant and $G=6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg }\mbox{s}^2}$ Newton's gravitational constant.
Hence with one parsec $1\mbox{pc}=3.0857×10^{16} \mbox{m}$,
$$\rho_c=\frac{3\cdot (67.8\cdot 10^3\cdot \frac{\mbox{m}/\mbox{s}}{3.0857×10^{22} \mbox{m}})^2}{8\pi\cdot 6.673848\cdot 10^{-11}\frac{\mbox{m}^3}{\mbox{kg}\mbox{s}^2}}=8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}.$$

Ordinary matter content is $4.9\%$, the universe has total density close to critical density, hence ordinary matter density is
$$0.049\cdot 8.635 \cdot 10^{-27}\frac{\mbox{kg}}{\mbox{m}^3}=4.23 \cdot 10^{-28}\frac{\mbox{kg}}{\mbox{m}^3}.$$
(The Hubble constant is not precisely known, so the calculated density depends a bit on the precise value.)

Density of water:
The density of water is about $10^3 \mbox{kg}/\mbox{m}^3$.
That's a factor $4.23 \cdot 10^{31}$ denser than the average universe.
The space needs to have been reduced by a factor $\sqrt[3]{4.23 \cdot 10^{31}}=3.48\cdot 10^{10}$ in each if the three spatial dimensions. That's the inverse of the scale factor, hence corresponds to a redshift of $z=3.48\cdot 10^{10}$.

Using the Cosmology Calculator on this website, the cosmological parameters $H_0 = 67.11$ km/s/Mpc (slightly different to the above value),
$\Omega_{\Lambda} = 0.6825$ provided by the Planck project, and setting $\Omega_M = 1- \Omega_{\Lambda} = 0.3175$ the age of the universe was 0.022 seconds at the redshift $z=3.48\cdot 10^{10}$, hence when it was of the average density of water.

Density of air at sea level:
The density of air at sea level is $1.225 \frac{\mbox{kg}}{\mbox{m}^3}$. The corresponding redshift is about $z=1.423\cdot 10^9$ yielding an age of the universe of 12.68 seconds.