Is the Casimir energy in CFT an observable? We know that if we transform a 2d conformal field theory from a plane to a cylinder with perimeter $L$, the ground state energy will be shifted by $$E = -\frac{c}{24L}$$ due to the Schwarzian derivative term in the transformation of stress energy tensor. 
This energy is the difference of a theory on a cylinder and the same theory on a plane. How can we compare the ground state energy of two theories on different spacetime? Therefore I would like to know is this energy a physical observable? And if not, why is it important?
 A: Of course, the free energy on the cylinder is not a measurable observable if you're given the theory on the infinite plane. But one can measure other observables which are proportional to the central charge, such as the two-point function of the stress-energy tensor.
There are situations where that expression is an observable. If you have a one-dimensional quantum system with periodic boundary conditions that flows to a (1+1)-dimensional CFT, then its ground state energy will generically be given by the formula
$$
E = E_1 L + E_0 - \frac{\pi v c}{6L} + \cdots, 
$$
where the higher-order terms are lower-order in $L$. (See below about the mismatch between our expressions.) Here, $E_1$, $E_0$, and $v$ are non-universal constants ($v$ is the velocity of excitations at low-energy, usually called the "speed of light" in a field theory textbook). Then it is possible to "measure" the central charge term. For example, say you do some Monte-Carlo simulations to obtain the velocity $v$ of excitations, and then numerically calculate the ground state energy for several (large) values of $L$ and match it to the above equation. This lets you determine $c$. 
In practice, it is much easier to extract central charge from the entanglement entropy. In particular, for an open one-dimensional quantum system, the entropy associated with tracing out half of the system is $S = (c/6) \log L$.
As a side-note, I think that what you are calling $L$ is really the radius of the cylinder, which is related to the perimeter by a factor of $2 \pi$. Finally, you are only considering the holomorphic sector, and above I'm everywhere considering also the antiholomorphic sector with an identical central charge. So that's why my expression is off by $4 \pi$ compared to yours.
