What volume of water will dilute an acidic solution back to neutral?
The very idea of diluting a solution back to (almost) neutral with pure water is unreasonable, precisely because it would require insane amounts of water. That being said, your derivation (though impractical) is not meaningless; it is just slightly wrong.

Like J. Ari said in the comments, there is an equilibrium involving $\ce{H+}$ and $\ce{OH-}$, and it will be shifted by the dissolved acid (that is, the concentrations will shift so the equilibrium constant may stay the same). Therefore you can't assume that water will always contain precisely $10^{-7}$M protons from autodissociation (plus maybe some extra from acid). It doesn't work that way.

How does it work, then?

$$\ce{[H+]\cdot[OH-]=10^{-14}}\\
\ce{[H+]-[OH-]=}c(\rm{HCl})$$
(since the acid is strong, and hence fully ionized even in not-so-diluted solutions)
$$\ce{[H+]=10^{-6.995}}\\
\ce{[OH-]=10^{-7.005}}\\
c\ce{(HCl)=10^{-6.995}-10^{-7.005}=10^{-6.995}(1-10^{-0.01})}\\
V_{total}=V_0\cdot{c_0\over c}\\
V_{added}=V_0\cdot\left({c_0\over c}-1\right)
=V_0\cdot\left({10^{6.995-pH_0}\over1-10^{-0.01}}-1\right)$$
So it is only 4.3 million times the original volume, which is about 2 times less than your estimate.

Not that it made much of a difference, really.