DC gain from an impulse response
The DC gain is simply the sum of filter taps or coefficients. This is the value of the frequency response at DC (i.e. $0\ \rm Hz$), or equivalently
$$
H(0) = \sum_{n = 0}^{N - 1}h[n]\tag{1}
$$
Because, for a digital FIR filter of length $N$ with impulse response given by Equation $(2)$
$$
\big\{h[n]\big\}, \quad\text{with}\quad 0\le n\le N -1\tag{2}
$$
the frequency response is the $z$-transform at the unit circle (i.e. $z = e^{j2\pi f}$), or in this case, equivalently
$$
H(e^{j\omega}) \equiv H(\omega) = \mathcal F \left\{h[n]\right\} \triangleq \sum_{n = 0}^{N - 1}h[n]e^{-j\omega n } = \sum_{n = 0}^{N - 1}h[n]e^{-j2\pi n f}\tag{3}
$$
Then you can think at DC (i.e. at $0\ \rm Hz$ or $f = 0$ or $\omega = 0$) to see the why of Equation $(1)$.
From DTFT to DFT
The DFT is the one practically computed in place of the DTFT. For finite-length signals (as is the case here), it provides the frequency-domain samples of the DTFT. Then $H[k]$, which is the DFT of the $h[n]$,  is computed for $N$-point as shown in Equation $(4)$
$$
H[k] = H(e^{j\omega}) \bigg\vert_{\omega = 2\pi k/N}\quad\text{with}\quad 0\leq k \leq N -  1\tag{4}
$$
i.e. $H[k]$ consists of equally-spaced (by $2\pi/N$) samples of $H(e^{j\omega})$.
Noting that
$$\omega \equiv 2\pi f \quad\text{with}\quad -\pi \leq \omega \leq \pi\quad\text{and}\quad -\frac 12 \leq f \leq \frac 12$$
With $\omega$ in [radians/sample] and $f$ in [cycles/sample].
In MATLAB
This can be done using MATLAB's freqz function as follows:
&gt;&gt; [h, w] = freqz([-3 0 -3 0], 1, [0 , pi/4])

h =

 -6.000000000000000 + 0.000000000000000i -3.000000000000000 + 3.000000000000000i


w =

                   0   0.785398163397448

&gt;&gt; 

Giving you the flexibility to specify at whatever angular frequencies $\omega$  (in example above the evaluation is at two frequencies $0$ and $\pi/4$ rad/samples) you would want to evaluate the frequency response, (Here we're interested in the value at $\omega = 0$).  The magnitude response with these two points is shown below

Note that the two points in magenta correspond exactly to
&gt;&gt; 20*log10(abs(h))

ans =

  15.563025007672874  12.552725051033063

&gt;&gt; 

The values at $\omega = 0$ and $\omega = \pi/4$ respectively.