UPDATE This version is substantially improved from the one posted at 8 AM.
I now think I can achieve $mathbb{Z}/p$ using $O( log p)$ vertices. I'm not trying to optimize constants at this time.
Let $B$ be a simplicial complex on the vertices $a$, $b$, $c$, $a'$, $b'$, $c'$ and $z_1$, $z_2$, ..., $z_{k-3}$, containing the edges $(a,b)$, $(b,c)$, $(c,a)$, $(a',b')$, $(b',c')$ and $(c',a')$ and such that $H^1(B) cong mathbb{Z}$ with generator $(a,b)+(b,c)+(c,a)$ and relation
$$2 {large (} (a,b)+(b,c)+(c,a) {large )} equiv (a',b') + (b',c') + (c',a').$$
I think I can do this with $k=6$ by taking damiano's construction with $p=2$ and adding three simplices to make the hexagon $(h_1, h_2, ldots, h_6)$ homologous to the triangle $(h_1, h_3, h_5)$.
Let $B^n$ be a simplicial complex with $3+nk$ vertices $a^i$, $b^i$, $c^i$, with $0 leq i leq n$, and $z^i_j$ with $0 leq i leq n-1$ and $1 leq j leq k-3$. Namely, we build $n$ copies of $B$, the $r$-th copy on the vertices $a^r$, $b^r$, $c^r$, $a^{r+1}$, $b^{r+1}$, $c^{r+1}$ and $z^r_1$, $z^r_2$, ..., $z^r_{k-3}$. Let $gamma_i$ be the cycle $(a^i,b^i) + (b^i, c^i) + (c^i, a^i)$.
Then $H^1(B^n) = mathbb{Z}$ with generator $gamma_0$ and relations
$$gamma_n equiv 2 gamma_{n-1} equiv cdots equiv 2^n gamma_0$$
Let $p = 2^{n_1} + 2^{n_2} + cdots + 2^{n_s}$.
Glue in an oriented surface $Sigma$ with boundary $gamma_{n_1} sqcup gamma_{n_2} sqcup cdots sqcup gamma_{n_s}$, genus $0$, and no internal vertices.
In the resulting space, $sum gamma_{n_i} equiv 0$ so $p gamma_0 equiv 0$, and no smaller multiple of $gamma_0$ is zero. We have use $3 + k log_2 p$ vertices. This is the same order of magnitude as Gabber's bound.
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