The example $G = \mathbb Z/2$ and $M = \mathbb{RP}^3$ works.
The inclusion $\mathbb Z/2\to\{\pm 1\}\subset\mathrm U(1)$ induces an isomorphism $H^3(B\mathbb Z/2, \mathbb Z/2)\to H^3(B\mathbb Z/2, \mathrm U(1))$, so we can pull the cocycles back to $\mathbb Z/2$ cohomology and evaluate on the $\mathbb Z/2$ fundamental class.
$\mathbb{RP}^3$ has two isomorphism classes of principal $\mathbb Z/2$-bundles, the trivial bundle $\varepsilon$ and the connected double cover $\xi$. Each determines a classifying map to $B\mathbb Z/2$, and hence a map in cohomology $\phi^*\colon H^*(B\mathbb Z/2; \mathbb Z/2)\cong \mathbb Z/2[\alpha]\to H^*(\mathbb{RP}^3; \mathbb Z/2)\cong \mathbb Z/2[x]/(x^4)$. For $\varepsilon$, this is the zero map; for $\xi$, this is the ring homomorphism induced by $\alpha\mapsto x$.
Since $H^3(B\mathbb Z/2;\mathbb Z/2)\cong\mathbb Z/2$, there are two cohomology classes of cocycles. Let $\omega$ be a coboundary, so that $\phi^*\omega = 0$ for all principal $\mathbb Z/2$-bundles, and hence the Dijkgraaf-Witten partition function is
$$\mathrm{DW}_{\mathbb Z/2, 0}(\mathbb{RP}^3) = \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \varepsilon} + \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \xi} = 1.$$
Let $\omega$ be a cocycle in the other cohomology class (in the notation above, $\alpha^3$). Then, $\varepsilon$ pulls it back to $0$, but $\xi$ pulls it back to $x^3$, and $\langle x^3, [\mathbb{RP}^3]\rangle = 1$, so
$$\mathrm{DW}_{\mathbb Z/2, \alpha^3}(\mathbb{RP}^3) = \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \varepsilon} + \underbrace{\frac{e^{i\pi(1)}}{2}}_{\text{from } \xi} = \frac 12 - \frac 12= 0.$$