Since we are collecting some constructive criteria in the comments, I thought it would be nice to find a way to constrain the possible determinants.

Let $S$ be the set $\{-1, 0,1\}$, but the criterion holds for any set which is invariant for sign change. Unless otherwise specified, all the vectors and coefficients are in $S$. Let's denote by $\textrm{Det}_S(n)$ the set of integers that are achieved as determinants of $n\times n$ matrices with coefficients in $S$. Then we have the following

**CRITERION** We have that $d \in \textrm{Det}_S(n+1), d \neq 0$ if and only if there exist $0 \neq d_1 , \ldots, d_{n+1} \in \textrm{Det}_S(n)$, $v_1, \ldots, v_{n+1} \in S^n$ and $c_1, \ldots c_{n+1} \in S$ such that
$$ \sum_i (-1)^i d_i v_i = 0$$
$$ \det(v_2| \ldots | v_n) = d_1 $$
$$ \sum_i (-1)^i d_i c_i = d$$

*Proof.* If $d \in \textrm{Det}_S(n+1)$, by Laplace expanding in the first row you get numbers $c_1, \ldots, c_{n+1} \in S$ and determinants $d_1, \ldots, d_{n+1} \in \textrm{Det}_S(n)$ such that
$$ d = \sum (-1)^{i} c_i d_i $$
Also, the $d_i$ satisfy the following condition: there exists $v_1, \ldots,v_{n+1}$ such that
$$ d_i = \det( v_1 | \ldots | \hat{v}_i | \ldots | v_{n+1} ) $$
Suppose WLOG that $d_1 \neq 0$. Otherwise, up to signs, you can rearrange the columns and get $d_1 \neq 0$.We want to show that the $n+1$ determinant conditions are equivalent to the first condition plus
$$ \sum_i (-1)^i d_i v_i = 0$$
If the determinant condition holds, since the determinant of $v_2, \ldots, v_{n+1}$ is not zero, they are a basis. Write
$$ v_1 = \sum_{i \ge 2} (-1)^i \lambda_i v_i $$
For some $\lambda_i$. Substituting into the $i$-th determinant condition we get
$$ d_i = \det ( (-1)^i \lambda_i v_i | v_2 | \ldots | \hat{v}_i | \ldots | v_{n+1} ) = (-1)^i \lambda_i \det (v_i | v_2 | \ldots | \hat{v}_i | \ldots | v_{n+1} ) = \lambda_i d_1 $$
From which we get $\lambda_i = d_i/d_1$. We used that the contributions along, for example, $v_2$, does not contribute to the determinant since $v_2$ is a column of the matrix. Subsstituiting into the expression for $v$ we get
$$ v_1 = \sum_{i \ge 2} (-1)^i \frac{d_i}{d_1} v_i$$
From which we get the desired constraint.

On the other hand, if the constraint hold and the first determinant is correct, the determinants containing $v_1$ can be computed using the expression for $v_1$ exactly as above, yielding the correct value $d_i$ for the $i$-th condition.

**EXAMPLES.**

**$\textrm{Det}_S(3)$**. Starting from $\textrm{Det}_S(2) = \{-2, \ldots, 2\}$, let's see for example that the maximum of $\textrm{Det}_S(3)$ is $4$.
Indeed, the tuple $222$ is impossible, since
$$ 2v_1-2v_2+2v_3 = 0$$
Implies
$$ v_1 = v_3-v_2$$
But the only pairs of vectors with determinant two are in $(\pm 1, \pm 1) $. This implies that $v_1$ has even entries, but the only possible even entry is zero. We conclude that $v_1-v_3 =0$, a contradiction.

The tuple $221$ is impossible too, since
$$ 2v_1-2v_2+v_3 = 0$$
Implies modulo 2 that $v_3=0$, a contradiction. 220 is possible and realized by $ (1, 1), (1, -1), (1, -1) $.

**$\textrm{Det}_S(4)$**. Going one step further, this explains why 15 does not show up in $\textrm{Det}(4) $. Indeed, the only possible tuple that would determine $15$ is $ 4443$, and this is not possible, since
$$ 4v_1-4v_2+4v_3-3v_4=0$$
Implies modulo 2 that $v_4=0$,which is a contradiction.

Similarly, $14$ is not possible, since the two candidate tuples are $4433$ and $4442$. The latter implies
$$ 4v_1-4v_2+4v_3-2v_4 = 0$$
Dividing by $2$ and analyzing modulo $2$ we get $v_4 = 0$, which is impossible. The tuple $4433$ gives
$$ 4v_1-4v_2+3v_3-3v_4 = 0$$
Modulo $4$ we have $v_3 = v_4$, which implies $v_3 = v_4$, a contradiction (since $4 = \det(v_2 | v_3 | v_4)=0$).

**Corollary criterion.** A nonzero candidate tuple cannot have only one odd number, nor only two equal numbers $a \neq 0 \pmod{m}$ numbers for $ m/ \textrm{gcd}(a,m) \ge 3$.

*Proof.* If there is just one odd number, modulo $2$ we get a zero vector, which is impossible. If there are only two equal numbers $a \neq 0 \pmod{m}$ we get modulo $m$ that
$$a v_i \equiv a v_j \pmod{m}$$
$$v_i \equiv v_j \pmod{\frac{m}{\textrm{gcd}(a,m)} }$$
which implies $v_i = v_j$, a contradiction.

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