I have tried to come up with a way, so I was wondering if it is correct, and if there are any other ways which work in this (or other) cases? Thanks.
Let $G = (\mathbb{Z}/2\mathbb{Z})^3$
Let $\phi$ be a homomorphism from $(\mathbb{Z}/2\mathbb{Z})^3 \to \mathbb{Z}/8\mathbb{Z}$. The order of the image of $\phi$ must divide $6$ and $8$, so it can only be $1$ or $2$. If the order is $1$, then $\phi$ must be the trivial homomorphism.
If the order of the image is $2$, then the order of the kernel is $4$, and the image must be $\{\overline{0}, \overline{4} \}.$ So all we have to do is find all subgroups of $(\mathbb{Z}/2\mathbb{Z})^3$ which have order $4$ (they will automatically be normal). Then, if $H$ is such a subgroup, we define $\phi$ to be $\overline{0}$ for all elements of that subgroup, it must be the case that $\phi$ to be $\overline{4}$ for all the other elements.
To see that any such map is a homomorphism, consider the following: Let $H$ be a subgroup of order $4$, and $\phi$ be the homomorphism constructed above. Then there is the natural projection homomorphism $\pi: G\to G/H.$ But $ H\cong \mathbb{Z}/2\mathbb{Z} \cong \{\overline{0}, \overline{4} \}$ (in $\mathbb{Z} / 8 \mathbb{Z}$), so any map of the type discussed above is a homomorphism from $(\mathbb{Z}/2\mathbb{Z})^3 \to \mathbb{Z}/8 \mathbb{Z}$.