Why do we deal only with large scale magnetic fields in astrophysics, and not electric fields?

Many astrophysical plasmas are well modeled as perfect conductors. Ideal MHD assumes this limit. As a result, there is no electric field in the fluid's rest frame. In other frames, we generally have $\vec{E} = -\vec{v} \times \vec{B}$, so there is an electric field. However, the perfect conductivity constraint means we don't have to model the electric field - if we evolve just the magnetic field (and the other properties of the fluid like its velocity and density), then we have the complete picture.

The natural followup question is, "Why can we assume infinite conductivity?" Most people's intuition about space is that it is mostly vacuum, and vacuum seems like as good an insulator as one can find. The thing about vacuum is that even though there are few charge carriers per unit volume, what charge carriers there are can proceed uninterrupted and respond to any electric field.

The book Physics of the Interstellar and Intergalactic Medium (Bruce Draine) gives some equations to quantify this. In eq. 35.48 it gives the conductivity of a pure hydrogen fully ionized plasma at temerature $T$ as $$ \sigma = 4.6\times10^{9}\ \mathrm{s}^{-1} \left(\frac{T}{100\ \mathrm{K}}\right)^{3/2} \left(\frac{30}{\log\Lambda}\right) $$ (CGS units), where kinetic effects and the Debye length are approximately captured by the Coulomb logarithm $$ \log\Lambda = 22.1 + \log\left(\left(\frac{E}{kT}\right) \left(\frac{T}{10^4\ \mathrm{K}}\right)^{3/2} \left(\frac{n_e}{\mathrm{cm}^{-3}}\right)^{-1}\right) $$ (eq. 2.17). Here $E$ is the particle kinetic energy, and $n_e$ is the number density of electrons.

To give some sense to these numbers, take a look at the conductivities in this table on Wikipedia. Copper has a conductivity of about $6.0\times10^7\ \mathrm{S/m} = 5.4\times10^{17}\ \mathrm{s}^{-1}$, so a $100\ \mathrm{K}$ hydrogen plasma is not nearly as conductive. However, drinking water has a conductivity of no more than $5\times10^{-2}\ \mathrm{S/m} = 4.5\times10^{8}\ \mathrm{s}^{-1}$, and air's conductivity is at most $8\times10^{-15}\ \mathrm{S/m} = 7\times10^{-5}\ \mathrm{s}^{-1}$. Thus astrophysical plasmas are not particularly insulating.

Bruce Draine's book also quotes a timescale for a magnetic field to decay over a lengthscale $L$: $$ \tau = 5\times10^{8}\ \mathrm{yr}\ \left(\frac{T}{100\ \mathrm{K}}\right)^{3/2} \left(\frac{30}{\log\Lambda}\right) \left(\frac{L}{\mathrm{AU}}\right)^2 $$ (eq. 35.49). Thus if the smallest length scales in your problem are at least $10\ \mathrm{AU}$ (and you are working around $100\ \mathrm{K}$), the magnetic field decay time due to the finite conductivity of the plasma (due in turn to e.g. ion collisions) is well over the current age of the universe. On smaller scales you may have to model such effects, and indeed many astrophysicists do just that.

The lack of the electric field in modeling plasmas stems from the Lorentz force, $$ \mathbf F=q\mathbf E+q\boldsymbol\beta\times\mathbf B $$ where $\boldsymbol\beta=\mathbf v/c$. For most astrophysical plasmas, the force is zero, so we have that $$ \mathbf E=-\boldsymbol\beta\times\mathbf B $$ So any time we see an electric field, we can simply replace it with the above cross product. This happens in Faraday's law: $$ \frac{\partial\mathbf B}{\partial t}=-\nabla\times\mathbf E=-\nabla\times\left(-\boldsymbol\beta\times\mathbf B\right)=\nabla\times\boldsymbol\beta\times\mathbf B $$

The justification for $\mathbf F=0$ is as Chris White says: we assume perfect conductivity.

This answer is complemantary to the answer by Kyle:

There exists a basic asymmetry: there are no magnetic monopoles of the number and dimensions that the electric monopoles exist ( there are theories with magnetic monopoles and people are looking for them but we are talking of masses much larger than electrons and quarks) .

Hand waving, (because I have not checked the math just extending the symmetry that would have to be imposed in maxwell's equations) , if there existed magnetic monopoles of the size/strength of electric ones ( electrons, quarks) then an alternate state of neutral (magnetically neutral) matter could have appeared, where the spill over electric dipole forces would dominate also over magnetically neutral matter the way magnetic forces do over electrically neutral matter, symmetrically. Then it would be relevant for astrophysics plasma observations, but this is science fiction.