Moving average filter magnitude attenuation
Question : How is the gain close to unity in the vicinity of 60Hz?
Answer   : It's not. Your $\omega_1$ mapping is not correct. Digital frequency $\omega \in [-\pi, \pi]$ in a filter response $H(e^{j\omega})$ is mapped to continuous frequency as follows :$$ f = \frac{\omega}{\pi} .\frac{f_s}{2}$$ $$\omega \in [-\pi, \pi] \rightarrow [-\frac{f_s}{2}, \frac{f_s}{2}]$$.

A Moving Average filter is a causal Low-Pass filter of which the cut-off frequency($\omega$) depends on $L_m$.

Intuition : It averages $L_m$ number of input samples, i.e. average of 1 current sample $x[n]$ and $L_m - 1$ past input samples. Averaging smoothens out variations from the input samples because the consecutive samples of output are changing very slowly. Why? Because, the variation in output sample can only be caused by the current input sample $x[n]$ which is small compared to sum of $L_m - 1$ past input samples. so, a large part of consecutive output samples remain approximately same. Hence, Moving average is a LPF and that is why useful to filter out noise with low computational complexity.

Now, since you have $|H(e^{j\omega})|$, so you can plot it against frequency and see where your desired band of 55Hz to 65Hz lies inside the Moving Average Filter response.
Figure 1 : Frequency response of the filter you have mentioned $L_m=30$, $F_s = 1440Hz$  and grey patch shows where the band 55Hz to 65Hz lies. It actually lies in the side lobe of MA response.

Figure 2 : Zoomed in Frequency response of the filter you have mentioned $L_m=30$, $F_s = 1440Hz$  and the patch shows where the band 55Hz to 65Hz lies.

You need to bring your interested band containing frequency from 55Hz to 65Hz inside the main Lobe and close to DC to get a less distorted 60Hz sinusoidal signal.
You can do 2 things to achieve your goal :

Decrease $L_m$ so that cut-off frequency of Moving Average(MA) filter moves away from DC and 60Hz sinusoid falls inside main lobe.

Increase sampling frequency $f_s$, so that input signal spectrum shrinks and as a result 60Hz comes inside the main and close to DC.



Option 1 : Decrease Filter length, $L_m = 10$, and $F_s = 1440Hz$

Option 2: Increase Sampling rate, $L_m = 30$, and $F_s = 30*1440Hz$


As you can see that the effect both methods is that 60Hz now falls inside Main Lobe and hence you will retain you r desired signal of 60Hz. But the 2 options are not same and $2^{nd}$ option of increasing sampling rate while keeping $L_m = 31$ will perform way better than decreasing $L_m$ to 10. Why?
Two main reasons:

Decreasing $L_m$ to a lower value will not have as good denoising effect as a higher $L_m$, since the averaging is done over a shorter length.

Increasing sampling rate and then low-pass filtering your spectrum will inherently increase SNR. You can think about how!