Why are the units of the Hubble constant $T^{-1}$ and not $\frac{L^3}{T}$? This post explains what the units of the Hubble Constant are, but not why they are what they are. The units are clear from the fact the constant relates distance and observed speed of galaxies.
But the Hubble constant is supposed to measure the rate at which space is expanding, right? 
So I would expect more appropriate units to be  $\frac{L^3}{T}$  rather than $T^{-1}$ .
Perhaps there is another metric that measures in these units?
 A: Typically, one thinks of Hubble's constant in terms of astronomical observation.  This begins with Hubble's empirical law:
$$v = H d$$
Where $v$ is the velocity derived from the redshift of a distant galaxy, and $d$ is the distance to it.  The classic observational units for $H$ are therefore $\frac{\rm m/s}{\rm Mpc}$.  Relativists hate carrying around more than one unit, though, so we convert the Megaparsecs to meters, and end up with inverse seconds (or meters, since $c$ gives us a way to convert back and forth).
More simply, we can also think of Hubble's constant in terms of the Robertson-Walker metric:
$$ds^{2} = -dt^{2} + a^{2}(t)d^{3}{\vec x}$$
Where $d^{3}{\vec x}$ is the 3-metric of a homogenous space.  Then, $a(t)$ tells us how "big" space is at the current time, and we can think of the rate of expansion as $H(t) = \frac{\dot a}{a}$${}^{1}$, which obviously has units of inverse time.
Note that $H$ is actually a function of $t$ and is not constant in time, except for some special cases.
EDIT:
Note, if what you care about is the time evolution of 3-volumes, you can see that these are proportional to $a^{3}V_{0}$.  A time derivative of this then gives you $\frac{dV}{dt} = 3{\dot a}a^{2}V_{0} = 3 H a^{3}V_{0}$, which has the units you want, but is still logically anterior to the ordinary Hubble constant.
${}^{1}$We divide by $a$ so that $H$ doesn't depend on our arbitrary choice of the scale of $a$, and simply reflects the relative rate of expansion.
A: Yes in fact there is another metric. That is the frequency:
$$f=\frac{1}{T}$$
That is what $T^{-1}$ is supposed to mean. Also the equation for the expansion velocity is $$v = H\cdot d$$, so in terms of units $s^{-1}$ makes perfect sense.
Though from my knowledge the unit of Hubbles constant is usually represented with: $\frac{\text{km}}{\text{Mpc} \cdot \text{s}}$.
However I do understand your point to deal with it in terms of volume expansion rate $V=\frac{L^3}{T}$, still let us think about the meaning of this equation.
The further two objects are, the faster they will be moving away from each other. The rate of expansion is constant, it is $H$, but due to the expansion of the space the phenomenon of distance comes in. The easiest way to imagine this is using the balloon analogy. Two dots on a ballon will have the distance of 10mm, two other dots a distance of 1000mm. If we know double the size of the balloon in a given time $t$ then the first pair will have a distance of 20mm, the second a distance of 2000mm. So one pair expanded with 10m/s, the other with a 1000m/s. The rate of expansion remained constant, the expansion speed depends on two given points due to the effects of geometry.
A: If $H$ would be defined as $H=x\frac{\frac {m} {sec}} m=x\frac 1 {sec}$, the value would, of course, be much smaller, but then we have to express $d$ not in $(Mpc)$ but in $(m)$ for the recess velocity (expressed in $\frac m {sec}$) to remain equal at distance d.
