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 $t=\infty$.
When there is a change in magnetic-flux, based on Faraday's law of induction & Lenz's law, we know that there is change in Potential Difference now, aside from the source V now we have a induced −V due to the change in magnetic-flux, and it opposes the current, why would it? I understood from lenz's law that it will, but not great detail as to why.
This comes from Maxwell's equations. Start with Faraday's Law:
$$ \nabla \times E = -\frac{\partial B}{\partial t}$$
Note the negative sign. If we integrate both sides by the area through which there is a magnetic field (think of the cross-section of a solenoid, for example), then we get:
$$ \int (\nabla \times E) \cdot dA = \oint E\cdot dl = - \frac{\partial \Phi}{\partial t}$$
The middle term follows from applying Stoke's theorem on the first term. The middle term is also the definition of voltage. The final definition is flux ($\Phi$) which is the product of the magnetic field and the cross-sectional area the field passes through, which is how one arrives at the third term.
Another thing, if the power-source can be increased, V can potentially increase to oppose the −V? And maintain A at the same value it was?
The 'opposite V' or back EMF acts to stabilize the current flowing. Suppose you increase V, and hence A, which increases the flux and hence slowly increases the back EMF. Then suppose that you stop increasing V. The back EMF will persist and slowly decay thereafter until the current stops changing (current reaches steady-state). If you were to instead keep increasing V, you'd keep increasing A, keep increasing the flux and keep increasing the back EMF (it would exacerbate the situation).
Best Answer
Well there are many questions in your question and I'll try to address them one by one.
As you may know,
$L* di/dt = V (t)$
one important thing to note about inductors is that unless there is an impulsive excitation(that is, the input is bounded), it's current can not make instantaneous jumps.
Therefore, when you apply the voltage, it will still have a 0 current, it will have a high slope, but 0 current nonetheless at $0^+$ where $0^+$ denotes the instant right after 0.
Ok, so using $L* di/dt = V (t)$ and basic first order circuit analysis coupled with a bit of differential equations, you get the current equation:
$i(t) = V/R *(1 - e^ {-tR/L})$
R is the total thevenin resistance in the circuit. This includes the internal resistance of the inductor. This is the Zero state response of the inductor and it does not take into account any form of initial condition, however it is not hard to implement that into the formula. As you can see, the current will only reach its final value as t -> infinity.
If you assume the inductor circuit has zero resistance, it would be a short circuit between the terminals of the voltage source. If you assume there is an output resistance of voltage source, you can separate it from the voltage source and use in the calculation as R.
Change in the magnetic flux is basically the self induced EMF. The differential equation basically points out to the fact that most things in the inductor occur in an exponential decay kind-of fashion. I hope this answers at least some of your questions. Let me know in the comments if anything is not clear.
Edit: Just to be clear * is multiplication in my answer, not convolution or something like that.