Why is current in a circuit constant if there is a constant electric field?

It is a analogic useful way to understand the resistive circuit, in the meaning that the electric expression $V = RI$ has the same form as the mechanical $F = kv$, in an environment where the drag force is proportional to the velocity.

In the case of a conductor, it is important to note that even whithout any applied field, the free electrons have momentum to all directions, and of different magnitudes. But there is no net flow without an E-field. The effect of the E-Field is to increase the fraction of electrons to one direction, and decrease the fraction to the opposite. The scattering with the lattice limits that net flow, and is responsible for the Joule effect of the electrical resistance.

Exactly as @Claudio Saspinski says. I was about to write an answer in those same terms when I saw his excellent answer. Think of the metal as an extraordinarily dense "fluid" in which particles moving (electrons) almost immediately acquire their limit velocity. The faster they want to go, the more resistance they find, in proportion to their own velocity. $\dot{x}=\textrm{const.}$ is a solution to the equation, $$ m\ddot{x}=mg-k\dot{x} $$

a potential difference created by a battery causes charge carriers to flow because of an electric force, which can be represented by having an electric field everywhere in the circuit.

A potential difference created by a battery only causes charge carriers to flow if there is a load connected to the battery terminals. If there is no load connected, there is a potential difference between the battery terminals but no flow of charge carriers.

My question is then this - if the charge carriers in a circuit are motivated by an electric force, then how can current be a constant value? I am assuming current to be a measurement of how much charge crosses a cross-sectional area perpendicular to the direction of flow, per time; furthermore, shouldn't the velocity components of the charge carriers in the direction of flow, which is proportional to current, increase because of the electric force that is motivating them in that direction?

The force of the electric field causes the charge carriers kinetic energy to flow, but collisions between the charge carriers and the atoms/molecules of electrical resistance in the circuit takes that kinetic energy away and dissipates it as heat. It is a continual process of the electric field giving the charges kinetic energy and the collisions taking the kinetic energy away that keeps the velocity of the charge carriers (i.e., the drift velocity) and thus the current constant. The end result is the electrical potential energy given the charge carriers by the battery is converted is either converted to heat in resistance, or stored as potential energy in the electric fields of capacitance or kinetic energy in the magnetic fields of inductance.

In thinking about this question, I was led to a possible explanation related to Ohm's Law - my textbook compares resistance to a frictional force, caused by the accumulation of collisions undergone by a charge carrier (which result in changes in velocity against flow direction).

And that is indeed a good mechanical analogy. Let's say you push a box at constant velocity on a floor with friction a distance d. The force you apply is analogous to the force of the electric field. The box is analogous to the charge. And the floor friction is analogous to the electrical resistance. The box moving at constant velocity is analogous to constant current. The work you do (Fxd) divided by the mass m is analogous is analogous to the voltage drop (work per unit charge).

Continuing, I would then assume that with a given voltage and given resistance, current changes so that the two exert equal effective forces on charge carriers - this is implying that a resistor would only be able to apply a specific amount of total force (work) at a time, and thus the charge per time must be changed to ensure that the work per charge is equivalent to voltage (this last bit seems the most questionable to me).

Having a little trouble following you. But you should understand that the potential difference, or voltage, between the terminals of a resistor is equal to the work required, per unit charge, or Joules/Coulomb, to move the charge between the terminals of the resistor. Then, since the current in the resistor is the charge per unit time, coulombs per second, moving across any point in the resistor, then the power dissipated in the resistor as heat is the voltage times the current, or Joules/Coulomb x Coulombs/Sec = Joules/sec = watts.

Also, when you said "the potential difference, or voltage, between the terminals of a resistor is equal to the work required, per unit charge...to move the charge between the terminals of the resistor" - to clarify - if this work per charge is a combination of both resistance and current, and current is charge per time, does this mean resistance is work per time?

First, what I said is the electrical engineering definition of potential difference, or voltage. In your example it represents the work required by the battery, per unit charge, to move the charge through the resistance.

Second, the definition of voltage is independent of current. There is a voltage across the terminals of a battery when it is not connected to anything, i.e., when there is no current. But the relationship between the current in and the voltage across a resistor of constant resistance is determined by Ohm’s law.

Third, resistance is not work per unit time. Work per unit time is the power dissipated in the resistor and equals $VI$ {Joules/Coul x Coul/sec = watts).

Hope this helps.