I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
1. Why self-inductance is not considered when solving Faraday's law problems
Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.
This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.
2. Why an EMF can ever produce a current in a circuit with non-zero self-inductance.
You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.
To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.
The magnetic field lines are in the same direction as the upwards velocity of the electrons, so the v × B term in the Lorentz force is 0.
Magnetic field lines are curved, so they cannot be everywhere in the same direction. Magnetic field in the metallic wire is not everywhere parallel to the velocity of the wire and produces electromotive intensity $\mathbf v\times \mathbf B$ circulating in the horizontal plane that is non-zero both below and above the center of the magnet. Near the center of the magnet, the induced intensity is low due to fact you mention, i.e. velocity is parallel to magnetic field.
Best Answer
It's not an ideal resistor - since ideal resistors have only resistance - and it's not an ideal inductor -since ideal inductors have only inductance.
If this were a loop of ideal conductor, which has zero resistance, a constant current could exist in the loop without an emf generating, time changing magnetic field linking the loop since there is no dissipation of energy.
However, when there is resistance in the loop, sustaining a current $I$ requires a non-zero emf since the resistance dissipates energy.
When the voltage across the resistance (given by Ohm's law) and the emf generated by the time changing magnetic field are of the same magnitude, the current is constant with time.