Question: Lévy processes are self-decomposable and independent on any non-overlapping interval, so how come the distribution of the process at time T,$\phi(T)$, which is the sum of N i.i.d with law $\phi(T/N)$ is not normally distributed ? 
I cannot find what am I missing here ?

Answer: 
The Lévy theorem states that the conditions that have to be met for $M(t)$ to be a Brownian motion (and hence be normally distributed):

* $M(t)$, for $t>0$, be a martingale relative to some filtration $F(t), t>0$.
* $M(0)= 0$
* $M(t)$ has continuous paths
* $[M,M](t) = t$ for all $t\geq0$;

So, to test each condition you simply differentiate your Lévy process (using Itô Calculus), change to the integrated form and you notice that at $t=0$ the stochastic integral takes the value zero, hence the expectation is always zero (at $t=0$). When you take the expectation of the stochastic integral you can isolate the following moment generating function:

$$\mathbb{E}\left[\mathrm{exp}\left(uM(t)\right)\right] = \mathrm{exp}\left(\frac{1}{2} u^2t\right)$$ 

which is the MGF for the normal distribution with zero mean and variance $t$. Therefore your $M(t)$ Lévy process follows the same distribution and you just showed that also an $M(t)$ process follows normality. 

If your Levy process does not satisfy above conditions but still follows the general definition of a Levy processes as stated here: 

<http://almostsure.wordpress.com/2010/11/23/levy-processes/>

then you can utilize the characteristic functions. Since the characteristic function of a convolution is the product of the characteristic functions of the densities involved, the central limit theorem has yet another restatement: the product of the characteristic functions of a number of density functions becomes close to the characteristic function of the normal density as the number of density functions increases without bound, under the conditions stated above. However, to state this more precisely, an appropriate scaling factor needs to be applied to the argument of the characteristic function. 

The following shows how those scaling factors and drift adjustments can be made: 

<http://www.stats.ox.ac.uk/~winkel/lp1.pdf>
A very good question. In other words you ask why the central limit theorem does not hold, right? A sum of iid should be somehow normal, right? Looking at the Levy-Kinchin representation we see the Gaussian part, which comes from increments of a continuous process, and the rest from the jumps. So one answer (which is not mathematically rigorous) is the presence of jumps. Another reason is that a Levy process can have infinite moments (also because of the jumps).

If we the process is continuous then it is Gaussian (if and only if). The jumps enrich the model but of course make if much more complicated.