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 trouble arises, I believe, because you're considering the field to be due to a current in a wire of zero thickness, so the flux density approaches infinity as you approach the wire, and this makes the flux integral blow up. If you consider current spread over a finite cross-sectional area of wire this problem goes away. There are other mathematical difficulties, of course, but they can be handled by approximation methods, and you'll find formulae for flux due to a circular loop on the internet.
Best Answer
Start with the vector potential $\textbf{B}=\textrm{curl}\textbf{A}$. For a magnetic dipole $\textbf{A}=\frac{\mu_0}{4\pi}\frac{\mathbf{\mathfrak{m}}\times\textbf{r}^0 }{|\textbf{r}|^2}$, this is also the asymptotic field at large distances away of any finite sized source. Now from Stokes' theorem the flux through any surface $S$ with boundary $\mathcal{C}=\partial S$ is $$\Phi = \int_{\mathcal{S}}\textbf{B}\cdot d\textbf{S}=\int_\mathcal{S} \textrm{curl} \textbf{A} \cdot d{\textbf{S}} \\ =\oint_{\mathcal{C}} \textbf{A} \cdot d{\textbf{c}}$$ Let $\mathcal{C}$ be a circle of arbitrary large radius, that is we are looking at the flux through an infinite sheet of surface, then the latter integral approaches zero for a dipole field because the perimeter is proportional to the radius while the integrand is proportional to the inverse squared radius. Then it also follows that the flux of an infinite sheet is also zero for any finite sourced magnetic field, in other words the flux threaded by any simple loop is as exactly equal to the (negative) flux threaded by the outside of that loop.