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 magnetic field is perpendicular to the current's direction, and the current is going normal to the loop, meaning that $B$ is parallel to the loop's plane. Because flux is the dot product of the magnetic field and the loop's normal "surface vector", and the angle is $90^\circ$, it is equal to $0$, thus not changing in time and not inducing an emf.
$$ \varepsilon = -\frac{d \Phi}{dt} = -\frac{d}{dt} (0) = 0 $$