why there is no 0 on the right side? (matlab, low pass filter)
Answer: Your derivation is absolutely correct and You will see 0 for both $\omega = -\pi$ and $\omega = \pi$, if you use fvtool() function of MATLAB instead of fft(). The reason is simple : FFT does not calculate values of $H(e^{j\omega})$ at continuous $\omega \in [-\pi, \pi]$, but it calculates DFT and for that the digital frequency resolution is $\omega = \frac{2\pi k}{N}$. So, for $N=6$, it will compute values of $H(e^{j\omega})$ at $\omega = 0, \frac{2\pi}{6}, \frac{4\pi}{6}, \frac{6\pi}{6}, \frac{8\pi}{6} \ and \ \frac{10\pi}{6}$. See that $\omega = \pi$ is occurring only at $k=3$, and hence you are seeing $0$ only once.

For checking this, you can take any symmetrical h, like $h = {1,2,4,4,2,1}$. 
Then use the following MATLAB code to plot its DTFT for $\omega \in [-\pi, \pi]$


  fvtool(h,1);


This will plot a Single sided DTFT from $\omega \in [0, \pi]$. You can then change from "Analysis" option to plot for frequency range $[-\pi, \pi]$.

You will get following plot for the example I have taken:


Check that the values are $0$ at both $\omega = -\pi$ and $\omega = \pi$.

And this will happen for any symmetric values, not any particular window function.