Permittivity. What does it permit? From the name of the concept, you would expect the permittivity of a medium to be the ability of the medium to permit something.
One may expect that, permittivity would be the ability of the medium to permit electrical fields through it. However, looking at Coulombs law and its derivatives,
$$
E = \frac{1}{4 \pi \varepsilon} \frac{Q}{r^2}
$$
one can tell that a high permittivity means a low electrical intensity (i.e., the electrical field not being permitted through) and viceversa. So it seems that in this scenario, should be named impermittivity 
Therefore, what does the permittivity of a medium permit?
 A: It permits the electric field to do something with the matter/medium. Forget the Coulomb’s force law for a moment, and consider the Gauss’ law. It states the following:
$$Q\propto\int_{\partial V}E\cdot\mathrm{d}s$$
where $Q$ is the total electric charge of the matter in the volume $V$, and $E$ the electric field. Now here the proportional constant is the permittivity of matter, denoted by $\varepsilon$, which is about how much the matter is permitting the electric field to do something effective with the matter.
Here, the word “permit” doesn’t mean how well the matter is permitting the field to exist throughout the volume, but how well the matter permits the field to itself.
For instance, if the matter is not permitting the field at all to itself, this would mean letting the field going through the matter with no interruption (i.e., vacuum, lowest possible permittivity). But if the matter is permitting the field to do something effective with the matter, then the field will lose some of its magnitude which results in lower flux at the boundary.
A: The permittivity of a medium is the ability of the medium to permit polarization of itself under the effect of a electric field.
When a medium is polarized due to such an electric field, another electric field is induced, where the field lines are opposite of the original one. As a result, the total electric field inside of the medium is diminished.
A: Consider a differential forms approach. E is the electric field intensity. It is a one-form represented by parallel planes that are more densely spaced at higher intensities. D is the electric flux density. It is a two-form represented by tubes that pierce the surfaces. These tubes are more densely compacted for larger D.
$D= \epsilon \star E$
$\epsilon$ depends on the medium.
$\star$ depends on the metric.
Maxwell's Equations relate $E,B,H,$ and $D$. $E$ and $D$ are not the same. This is usually glossed over in introductory textbooks, but when  experiments were done that verified Maxwell's equations they needed to keep track of this difference. The equations relate field intensities to a flux densities. And simplifying units obscure that.
