What was the age of the universe when the average density was one atmosphere? How long did it take after the Big Bang for the average density of the universe to reach one Earth atmosphere at sea level? How about the density of water? Is there a chart of densities that relate to everyday objects (so easier to understand)?
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.