Calculation frequency response of digital filter with known structure
For the given system you can write down the input-output relation as

$$y[k]=\frac14\left(x[k]+2x[k-1]+x[k-2]\right)\tag{1}$$

because $T$ (or $z^{-1}$) denotes a delay element, which delays its input by one sample interval.
The $\mathcal{Z}$-transform of (1) is (assuming zero initial conditions)

$$Y(z)=\frac14\left(X(z)+2X(z)z^{-1}+X(z)z^{-2}\right)=\frac{X(z)}{4}\left(1+2z^{-1}+z^{-2}\right)\tag{2}$$

From (2) you get the system's transfer function

$$H(z)=\frac{Y(z)}{X(z)}=\frac14\left(1+2z^{-1}+z^{-2}\right)\tag{3}$$

Since the system is stable (any FIR filter is), the frequency response can be obtained by evaluating the transfer function on the unit circle $z=e^{j\omega}$:

$$H(e^{j\omega})=\frac14\left(1+2e^{-j\omega}+e^{-2j\omega}\right)\tag{4}$$

You also could have arrived at (4) by writing down the impulse response from (1)

$$h[k]=\frac14\left(\delta[k]+2\delta[k-1]+\delta[k-2]\right)\tag{5}$$

and taking the (discrete-time) Fourier transform, which also results in (4).

Finally, a few notes concerning misconceptions in your question:


$\mathcal{L}$ usually denotes the Laplace transform, which is defined for continuous functions. $X(z)$ is the $\mathcal{Z}$-transform of the sequence $x[k]$, which can be written as $X(z)=\mathcal{Z}\{x[k]\}$.
The frequency response of a linear time-invariant (LTI) system is the Fourier transform of the system's impulse response: $H(e^{j\omega})=\mathcal{F}\{h[k]\}$, not the Fourier transform of $H(z)$ (which you seem to think, judging from the last equation in your question).