By definition, Hodge dual operator is defined as follows:
Let $\sigma = (i_1, i_2, ..., i_n)$ be a permutation of (1, 2, ..., n), then for any $k \in {0,1,...,n}$, the Hodge dual of the corresponding elementary k-form is:
$*(dx_{i_1} \wedge dx_{i_2} \wedge ... \wedge dx_{i_n}) = sgn(\sigma)\epsilon_{i_1} \epsilon_{i_2} ...\epsilon_{i_k}dx_{i_{k+1}} \wedge dx_{i_{k+2}} \wedge ... \wedge dx_{i_n}$, where $sgn(\sigma)$ is the sign of $\sigma$ and $(\epsilon_1, \epsilon_2, ..., \epsilon_n) \in \{+1, -1\}^n$ is the signature of the metric tensor.
If we pick the Minkowski spacetime with signature $"+---"$, then $n=4$ and $\epsilon_t = -\epsilon_x=-\epsilon_y=-\epsilon_z=1$.
Now I am trying to compute $*(dy)$.
Since $(y,t,x,z)$ is an even permutation of $(t,x,y,z)$, the signature is positive and so: $*(dy)= 1 \cdot (-1) dt \wedge dx \wedge dz = - dt \wedge dx \wedge dz$
The answer shows $*(dy)= dt \wedge dx \wedge dz$, which differs by a sign with my result and this may seem stupid but I cannot find where I went wrong.
You are correct, it should have a negative sign (see https://en.wikipedia.org/wiki/Hodge_star_operator#Four_dimensions). Possible reasons for the discrepancy with the given answer are that it's a misprint, that the answer uses the convention $(-+++)$ for Minkowski spacetime, or that the problem is actually asking about a different 4-dimensional space (like $\mathbb{R}^4$ with the standard metric).