In the book by P.G. de Gennes Superconductivity of Metals and Alloys, it's written
$N(0)V < 0.3$.
Also written :
Lead and mercury are two notable exceptions with low $\Theta_{D} \left( =\hslash \omega_{D}/k_{B} \right) $, giving, respectively, $N(0)V=0.39$ and $N(0)V=0.35$.
More details on p.112, P.G. de Gennes Superconductivity of Metals and Alloys, Westview (1999). The first edition dated back to 1966. To my knowledge there is no change between the editions.
See also the table 4-1 on p.125 of the same book for several specific values for the pure metals. This table is reproduced below for commodity.
$$\begin{array}{cccc}
\mbox{Metal} & \Theta_{D}\left(\mbox{K}\right) & T_{c}\left(\mbox{K}\right) & N\left(0\right)V\\
\mbox{Zn} & 235 & 0.9 & 0.18\\
\mbox{Cd} & 164 & 0.56 & 0.18\\
\mbox{Hg} & 70 & 4.16 & 0.35\\
\mbox{Al} & 365 & 1.2 & 0.18\\
\mbox{In} & 109 & 3.4 & 0.29\\
\mbox{Tl} & 100 & 2.4 & 0.27\\
\mbox{Sn} & 195 & 3.75 & 0.25\\
\mbox{Pb} & 96 & 7.22 & 0.39
\end{array}$$
Post-scriptum:
There are (surprisingly !) nothing about this question on the book by J.R. Schrieffer, Superconductivity, Benjamin (1964). There is not even a discussion on the gap equation as far as I can check... There is a repetition of the Gennes data on the book by A.I. Fetter and J.D. Walecka, Quantum theory of many-particle systems, Dover Publications (2003, first edition 1971), p.448. But this table is less complete.
There is also no discussion about the numerical value in the original paper by BCS [Bardeen, J., Cooper, L. N., & Schrieffer, J. R. ; Theory of Superconductivity. Physical Review, 108, 1175–1204 (1957). http://dx.doi.org/10.1103/PhysRev.108.1175 -> free to read on the APS website], but there is a possibly interesting expression (Eq.(2.40), written below in your notation / in the original BCS paper, the gap is written $\varepsilon_{0}$):
$$\Delta\left(T=0\right)=\frac{\hslash\omega_{D}}{\sinh \frac{1}{N(0)V}}$$
which might be of help for calculating the critical line $\Delta(T)$.
First of all, at zero temperature not all of the electrons in the BCS ground sate (GS) form Cooper pairs. One way to think about this is that deep inside the Fermi surface, there is no available states for scattering events, including the phonon-mediated ones, and thus we wouldn't have the effective attraction to form the Cooper pairs.
This can also be seen from the coefficients $u_k$ and $v_k$ in the postulated BCS GS given in the original post. When $k$ is way below the Fermi surface, then $|\xi_k|>>\Delta$, with $\xi_k<0$. In this case, $|u_k|\to 0$ and $v_k\to 1$, i.e. the states far below the Fermi surface are almost fully occupied by electrons created by $c^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}$. These electrons are created in pairs, but they are NOT Cooper pairs!!! The pair creation operator does not tell us anything regarding whether the pair is a Cooper or just a normal pair. Only when it is close to the Fermi surface, and thus possible to have phonon-mediated scattering, is the Cooper pair creation possible.
In summary, the BCS GS looks like the following: deep down inside the Fermi surface, it's just normal electrons. Close to the Fermi surface, it's mostly a superposition $\left(u_k+v_kc^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}\right)|0\rangle$ of empty state, with amplitude $u_k$, and occupied Cooper pairs, with amplitude $v_k$.
EDIT:
After I submitted the above answer, I realized that I missed out the other part of your question, which is about the excitations of the BCS GS and here it goes:
You mentioned that "the excitations of the BCS state must be created or destroyed in pairs". This is not correct. The excitations of the BCS GS are fundamentally different from Cooper pairs or the normal electrons. Instead, it is the quasi-particle created by
$$\gamma^{\dagger}_{p\uparrow}=u^*_pc^{\dagger}_{p\uparrow}-v^*_pc_{-p\downarrow}$$
Not surprisingly, this is just the Bogolyubov transformation used to diagonalize the BCS mean field hamiltonian and this why it is the legitimate excitations of the GS. The corresponding eigenvalue is exactly
$$E_k=\sqrt{\xi_k^2+\Delta^2}$$
This is where the gap $\Delta$ to the BCS excitation comes from. This kind of excitation is slightly less intuitive compared to objects like single electrons or Cooper pairs. It has spin-1/2, but it doesn't carry well-defined charge, whereas a Cooper pair has spin 0 and charge $2e$.
Also, You can easily check that
$$\gamma_{p\uparrow}|BCS\rangle=\gamma_{-p\downarrow}|BCS\rangle=0$$
which makes perfect sense since the annihilation operator for the excitation annihilates the GS.
Best Answer
The reason is that those are two different calculations with two different integrals but the hypothesis that link them is made before arriving at the final form.
In the Cooper Problem the hypothesis on the potential states that it is non-zeri and attractive when the energies of the electrons are in $[E_F ; E_F+ \hbar \omega_D ]$ and comes from the rigid Fermi sphere hypothesis plus the phonon induced attractive potential.
In the Bogoliubov approach the same hypothesis on the potential is made but in this approach you get an equation for the energy gap parameter that is taken non-zeri for $|\epsilon_k|< \hbar \omega_D$ because you don't have the rigid Fermi sphere hypothesis since all the electrons participate.
Now, this being said, one could also state that the interval are made such that the exponentials give the same results, indeed $\hbar \omega_D$ gives you an order of magnitude of the interaction and it's meant to give results that are qualitatively correct more than quantitatively. Think about the fact that those two approaches are talking about systems that are also different, why would the interaction be the same for a system with a rigid Fermi sphere and another where this assumption isn't made?
I think that you must watch at those results qualitatively more than quantitatively and, if you want, you can say that the intervals are taken such that they give the same results