Dark energy exerts a repulsive gravitational influence.
Dark energy is not normally considered to be itself a force. Instead it is a substance or form of energy that is uniformly distributed throughout the universe, and (like everything else) it exerts a gravitational influence, but its gravitational influence is repulsive. This verbage matches how dark energy is treated mathematically in general relativity. Whereas matter is modeled as a (nearly) pressureless perfect fluid and radiation is modeled as a fluid with pressure $p=\rho/3$ (where $\rho$ is the energy density and I take $c=1$), dark energy is a fluid with $p=-\rho$. Due to how both pressure and energy density contribute to gravity, a fluid with $p<-\rho/3$ exerts gravitational repulsion inside it.
At what distance does dark energy's gravitational repulsion balance a mass's gravitational attraction?
For a mass $M$ and uniformly distributed dark energy with density $\rho_\Lambda$, the gravitational attraction and repulsion balance when the distance $r$ from the mass is such that the average enclosed mass density is twice the dark energy density, i.e. $M/(4\pi r^3/3)=2\rho_\Lambda$. This leads to
$$r = \left(\frac{3}{8\pi}\frac{M}{\rho_\Lambda}\right)^{1/3}. \tag{1}$$
The density comparison comes from the second Friedmann equation, according to which the gravitational acceleration of a spherical shell is proportional to $\rho+3p$ (taking $c=1$), where $\rho$ and $p$ are the enclosed density and pressure, respectively. Matter has zero pressure while dark energy has $p=-\rho_\Lambda$, so the gravitational acceleration is proportional to $\rho_m-2\rho_\Lambda$, where $\rho_m$ is the matter density.
If dark energy is a cosmological constant, its value is $\Lambda=8\pi G\rho_\Lambda$, so an alternative expression is
$$r = \left(\frac{3GM}{\Lambda}\right)^{1/3}.\tag{2}$$
This can alternatively be seen to follow directly from the Newtonian limit of de Sitter space. In the Newtonian limit, the gravitational acceleration induced by dark energy at position $\vec r$ with respect to any freely falling observer (at $\vec r=0$) is
$$\ddot{\vec{r}}=\frac{\Lambda}{3}\vec r=\frac{8\pi G}{3}\rho_\Lambda \vec r.$$
In our universe
The density of dark energy is about $9\times 10^{10}$ M$_\odot$ Mpc$^{-3}$, so from equation (1), the forces of the matter and dark energy balance when
$$r\simeq 1.1~\mathrm{Mpc}\left(\frac{M}{10^{12}~\mathrm{M}_\odot}\right)^{1/3}.\tag{3}$$
Notice that $10^{12}~\mathrm{M}_\odot$ is about the mass of the Milky Way, and it's within a factor of a few of the mass of the Local Group. Thus, in the asymptotic future (assuming dark energy is a cosmological constant), only a sphere about 1 Mpc in radius is expected to remain bound to us. For reference, that's comparable to the present distance to the Andromeda galaxy (which is about 0.75 Mpc away).
