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 $t=\infty$.
Generally, it isn't the case that the (net, total) magnetic flux threading the conducing loop is constant.
Note that the quote uses the word opposes which you have evidently taken to mean nullify. But in fact, if the solution requires an emf, then there must be a non-constant flux threading the loop.
When the conductive material forming the loop has non-zero resistivity, the solution must satisfy
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt}$$
and
$$\mathscr E = R i$$
where $R$ is the resistance 'round the loop and $i$ is the current.
But
$$\Phi = \Phi_\textrm{ext} + \Phi_i = \Phi_\textrm{ext} + L i$$
where $L$ is the inductance of the conducting loop.
Combining the above yields
$$R i + L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
For example, consider the case that $\Phi_\textrm{ext} = \Phi_0$ is constant then the solution is
$$i(t) = I_0e^{-\frac{R}{L}t}$$
$$\Phi(t) = LI_0e^{-\frac{R}{L}t} + \Phi_0$$
$$\mathscr E(t) = -\frac{\mathrm d\Phi}{\mathrm dt} = -\left(-\frac{R}{L} \right)LI_0e^{-\frac{R}{L}t} =Ri(t)$$
So, this is all consistent and notice that the flux is not generally constant though the external flux is.
Now, consider the case that $\Phi_\textrm{ext}$ is increasing linearly with time
$$\Phi_\textrm{ext} = \Phi_0 + \phi t$$
then the particular solution is
$$i(t) = -\frac{\phi}{R}$$
$$\Phi(t) = \left(\Phi_0 -\frac{L}{R}\phi\right) + \phi t$$
$$\mathscr E = -\phi = Ri $$
Again, this is consistent and the flux is not generally constant. However, note that this result depends on $R$ being non-zero.
For the $R = 0$ case, we see that
$$L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
or
$$\Phi_i = - \Phi_\textrm{ext} + \mathrm{constant}$$
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt} = 0$$
thus we conclude that, for a perfectly conducting loop, the magnetic flux threading the loop is constant.
Best Answer
The induced EMF depends upon the rate of change of the total flux, and not just the externally applied flux.
If a current present in a wire, that current induces a magnetic field. If the current is varying the magnetic field induced by the current will vary. That varying magnetic field will in turn induce an EMF. That is the basis for the phenomenon known as self-inductance.
When a coil of wire is used as an inductor in an electric circuit, it's principle of operation is self-inductance. When a voltage is applied across the terminals of the (stand-alone) coil, current begins to flow. However, that current creates a changing magnetic field, which causes an EMF to be created which opposes the originally applied voltage. The result is that the current in the coil/inductor rises over time rather than suddenly.
So, once again, the EMF induced in a coil depends not only on an externally applied changing magnetic field, but upon the total magnetic field, including that created by the changing electric current in the coil.