I will make a small calculation here, but please proceed to the results if you may like to.
Calculation
Stars are spherical and static, so metric near their surface (photosphere) and outside on is Schwarzschild. Hence time-time metric component on the surface is:
$$g_{44}=1-\dfrac{R_{grav,*}}{R_*}$$,
where $R_*$ is the radius of the star and $R_{grav,*}$ is its gravitational radius.
Then, if the velocity of the star is much smaller than the speed of light, gravitational redshift in the lowest order does not depend on this velocity. Therefore, the emitting star can be assumed at rest.
The light from the star propagates along isotropic geodesic in Schwarzschild metric. The geodesic is described by lagrangian:
$$ \mathcal{L} = \dfrac{1}{2}g_{\mu\nu}k^{\mu}k^{\nu}$$,
where $k^\mu=(\vec{k},\omega/c)$ is the 4-vector of the light wave and $\omega$ is the frequency of light. Since the metric is static $\dfrac{d\mathcal{L}}{dk^4}=g_{\mu 4} k^4=g_{44}k^4=\textrm{const}$. Therefore:
$$(1-\dfrac{R_{grav,*}}{R_*})\omega = \textrm{const}$$
for the light as it travels towards us. So:
$$\omega_{obs}=\omega_{emitted}(1-\dfrac{R_{grav,*}}{R_*})\Longleftrightarrow \lambda_{obs}=\dfrac{\lambda_{emitted}}{(1-\dfrac{R_{grav,*}}{R_*})}$$,
where $\lambda$ is the wavelength.
Redshift is simply $z=\dfrac{\lambda_{obs}-\lambda_{emitted}}{\lambda_{emitted}}$. Assuming $z\ll 1$ one has a simple formula:
$$
z_0=\dfrac{R_{grav,*}}{R_*}
$$
If $z_0$ turns out to be comparable to unity, one should compute
$$
z=\dfrac{1}{1-z_0}-1,
$$
which then gives the correct value of redshift. Note that redshift does not depend on $\lambda$.
Nice numerical forms for this would come from $R_{grav,*}=2.95\textrm{km}\dfrac{M_*}{M_\odot}$:
$$z_0=0.295\dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot}\Longleftrightarrow z_0=4.24\cdot 10^{-6}\dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}$$
It is also nice to express the redshift in $\textrm{km}/\textrm{s}$:
$$
z_0=8.84\cdot 10^4 \dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot} \textrm{km/s}\Longleftrightarrow z_0=1.27 \dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}\textrm{km/s}
$$
Summary and discussion
In summary, when redshift $z$ is small, it is approximated by $z_0$, the numerical expressions for which are given just above. If $z_0$ turns out to be not small, one can calculate $z=\dfrac{1}{1-z_0}-1$, which then gives correct redshift.
Stars
One can see that:
- For normal stars like the Sun ($R_*\sim R_\odot, M_*\sim M_\odot$) the redshift is of order $1\textrm{km/s}$. It is almost important since stars in solar neighbourhood move normaly at the velocity of a few tens of $\textrm{km/s}$
- For white dwarfs ($R_*\sim 10^4 \textrm{km}, M_*\sim M_\odot$) redshift is a few times $100 \textrm{km/s}$ and gets very important when doing proper spectroscopy. Therefore one typically accounts for it.
- For neutron stars ($R_*\sim 10 \textrm{km}, M_*\sim M_\odot$) redshift is very important $z\sim 0.4$, but neutron stars are general relativistic objects anyway, so it would be expected beforehand.
So, in summary, when measuring light from individual stars one does have to account for gravitational redshifts to get accurate results, and particularly so when studying white dwarfs.
Groups of objects
Now, the same formulae are correct to an order of magnitude when applied for larger volumes of space, with $R_*$ and $M_*$ now meaning the size of the volume and the mass inside it. However, as typical interstellar distances are of order of parsec and $\textrm{pc}=3\cdot10^{13}\textrm{km}$, resulting $z$ will be very small even for such dense groups as globular clusters ($z$ is of order $10^{-8}$ in this case). So, groups of objects do not affect the redshift.
Cosmological overdensities
Nevertheless, cosmological scale underdensities of order few $100\textrm{Mpc}$ in size may affect the apparent redshift of distant objects, as we would be inside the underdensity. However, such an underdensity would have to be significantly symmetric around us in order to explain the lack of corresponding anisotropy in cosmic microwave background. Therefore, it is considered unlikely.