What explains the high specularity of metals?
Warning: I am not a physicist.

As Dan Hulme already explained, light can't travel through metals, so dealing with IOR is a lot more... complex. I will answer why that happens and how to calculate the reflection coefficient.

Explanation: Metals are filled with free electrons. Those electrons react to external fields and reposition until electrostatic equilibrium is met (the electric field is zero inside a conductor in electrostatic equilibrium). When electromagnetic waves hit a metallic surface, the free electrons move until the field that they create cancels the field of the incoming wave. Those electrons grouped together radiate a wave going out nearly the same as the one that hit the surface (i.e. with very low attenuation). How much is attenuated depends on the material properties. 

From this explanation it is clear that conductivity is a key part of the high reflection coefficient on metals.

Math-wise, what you are missing is the complex index of refraction. On good conductors, such as metals, the complex term of the IOR is relevant and key for explaining this phenomena.

Practically, in rendering, achieving good metal parameters is more visual based. Artists adjust to their preference until it looks believable. Often you see a metalness parameter with specific handling for materials marked as metal.

Involved answer:

The complex index of refraction can be seen if we use Ohm's Law $J = \sigma \vec{E}$, which holds for conductors, on the Ampère-Maxwell equation using sinusoidal waves $\vec{E} = e^{i\omega t}$:

$$
\vec{\nabla} \times \vec{H} = \sigma\vec{E} + \frac{\partial \vec{D}}{\partial t} = \sigma \vec{E} + i\omega \epsilon \vec{E}
$$
$$
 = i\omega \left( \epsilon - i \frac{\sigma}{\omega} \right)\vec{E} = i \omega \epsilon_m\vec{E} 
$$

Note how we can interpret that whole term as a complex permittitivity $\epsilon_m$ and that $\sigma$ is the conductivity of the material. 

This affects the IOR, as its definition is given by:

$$
  n' = \sqrt{\frac{\epsilon_m}{\epsilon_0}} = \sqrt{\frac{\left(\epsilon - i \sigma / \omega\right)}{\epsilon_0}} = n_{\text{real}} + in_{\text{img}}
$$

This shows how $n'$ can be complex. Also, note how very good conductors have a relevant complex term, as $\sigma \gg \epsilon_0 \omega$. Since it would take a lot, I will skip some steps with a reference, page 27: it can be shown that, since $\sigma \gg \epsilon_0\omega$, (we are dealing with $\omega$ of the visible spectrum):
$$
  n_{\text{real}} \approx n_{\text{img}}
$$

and reflection from metals with normal incidence, from a medium with IOR $n$, given that $n' \gg n$:

$$
  R = \frac{(n_{\text{real}} - n)^2 + n_{\text{img}}^2}{(n_{\text{real}} + n)^2 + n_{\text{img}}^2} \approx 1
$$

Agreeing that a good conductor is, in general, a good reflector.

The famous Introduction to Electrodynamics from Griffiths, pages 392-398, explains this and a lot more in a similar fashion.