Other answers address the question of gravity, so I'll just expand on the atmosphere topic.
The Problem
The biggest physics issue with your proposed world is the atmosphere.
A star is formed when enough gas is present that the gravity from all of the gas is enough to collapse it down into a dense hot sphere. It would thus collapse any atmosphere around it into itself. First you may think, well maybe the star only collapses the heavy elements and the light ones could still form an atmosphere. However, most of a star is hydrogen, the lightest element. 473
So this is not true.
The density and pressure of a gas grows exponentially towards a the source of gravity, so it is impossible to have an extremely thick atmosphere without reaching pressures where things form plasmas. The exponential growth is because each layer of gas must have a pressure that can support all of the weight above it.
Even if the star initially didn't have enough gravity by itself to suck down a giant atmosphere, even a low density atmosphere would have more mass and thus gravity than a typical star.
The Solution
If you want to have a giant atmosphere, consider having it on the outside of a shell. So then you'd have a star, a large vacuum, a (possibly transparent (maybe even diamond)) inner shell, an atmosphere, and an optional outer shell.
With this arrangement your gravity in the atmosphere would be:
$$g=\frac{G\,(M_{star}+M_{shell}+ M_{atm})}{r^2}$$
Where $r$ is the distance to the center of the star, $g$ is your gravitational acceleration, $G$ is the universal constant of gravitation, and the $M_{star}$, $M_{shell}$, and $M_{atm}$ are the masses of the star, inner shell, and portion of the atmosphere closer to the star than $r$ respectively.
So now let's take a look at the equations for the atmosphere to see if we can come up with some numbers that will fit your criteria:
First the specific ideal gas law relating temperature $T$, density $\rho$, and pressure $P$:
$$\rho=\frac{P}{RT}$$
Where $R$ is the specific gas constant (for air = $286.9\frac{J}{kg\,K}$)
Let's say the temperature is constant to simplify our analysis, and keep our creatures comfortable.
The change in mass of atmosphere closer than $r$ as $r$ increases will just be the surface area of the sphere of size $r$ times the density at that $r$:
$$\frac{d\,M_{atm}}{dr}=4\,\pi\,r^2\,\rho=\frac{4\,\pi}{RT}\,r^2\,P$$
Then since the pressure of the atmosphere must support the weight of the gas above it, the rate of change is also related to the density:
$$\frac{dP}{dr}=\rho\,g=\frac{P}{RT}\frac{G\,(M_{star}+M_{shell}+ M_{atm})}{r^2}$$
To simplify things a little let's change our variables:
$$M=M_{star}+M_{shell}+ M_{atm}$$
$$\frac{dM}{dr}=\frac{dM_{atm}}{dr}$$
Now we almost have enough information to do a numerical integration; we just need our initial values, and our constants. So let's try:
$$T=25^\circ C$$
$$M_0=M_{sol}= 10^{30} kg$$
$$P_0=1 atm = 10^5 Pa $$
$$r_0 = 1 AU = 1.5\times 10^{11} m$$
Integrating numerically we can get a plot of pressure vs altitude:
As you can see, the pressure drops off to less than three quarters of the initial pressure (enough to cause altitude sickness) by about 5000 km. Certainly a thicker breathable atmosphere than the measly 2.4 km that earth has, but let's see if we can do better.
By increasing our starting radius and decreasing the mass of our star and shell we can decrease the rate of pressure drop off, so let's look at a start with near the minimum mass to still be a red dwarf, about a tenth of our sun, and let's start out 100 times as far away:
For this system it looks like the breathable atmosphere would extend out to 7000 km: not much of an improvement for the extremes it took to get there. 24