Here is a proposition and a conjecture, which together would establish an algorithm for determining whether a multilinear $f$ has a root.
Proposition
Let $c$ be the coefficient of $x_1x_2\cdots x_t$ in a multilinear $f(x_1,\ldots,x_t)$.
Let $k$ be the sum of the absolute values of the coefficients in $f$.
If $c>0$, then $f$ has roots iff it has roots where some $x_i$ has $|x_i|<k/c$.
Proof: Suppose all the $x_i$ have $|x_i|\ge k/c$. Let $P = x_1x_2\cdots x_t$. We rewrite the equation $f(x_1,\ldots,x_t)=0$ so the left side has the term with $P$, and the right side has everything else. The left side is $cP$. On the right side, each product of variables is at most $P/(k/c)$, since each of those products is missing at least one factor that goes into $P$. Taking absolute values gives $c|P| \le (k-c)|P|/(k/c)$, which is impossible.
Conjecture
Let $c$ be the coefficient of $x_1x_2\cdots x_t$ in a multilinear $f(x_1,\ldots,x_t)$.
If $c=0$, then $f$ has roots iff its constant coefficient is divisible by the gcd of the non-constant coefficients.
Comments: Will Sawin's answer offers more comments on this. Here I record some of the more elementary cases: The conjecture clearly holds if $f$ is linear, e.g. $f(x,y,z)=6x + 10y+ 15z + 7$ or $6x+10y+30z+7$. So in particular it holds for $f(x)$ and $f(x,y)$. I have verified that the conjecture holds for $f(x,y,z)$ mod 5, mod 7, mod 8 and mod 9, where a non-trivial example may look like $f(x,y,z)=xy+yz+xz+x-5$. And for $f(w,x,y,z)$, I can either prove the conjecture outright or reduce it to the three-variable case so long as $f$ has a coefficient of 0 for one of $wxy$, $wxz$, $wyz$ or $xyz$.
Algorithm conditional on the above
If $t=1$ it is trivial to determine if $f$ has a root.
If $t>1$ and $c=0$, we can determine whether $f$ has a root according to the above conjecture.
If $t>1$ and $c\neq 0$, let $d=\lfloor k/|c|\rfloor$. Then we can determine whether $f$ has a root by substituting the integers in $[-d,d]$ for each variable. Specifically, we test whether $f(-d,x_2,\ldots,x_t)$ has a root, and whether $f(-d+1,x_2,\ldots,x_t)$ has a root, making all possible substitutions until testing whether $f(x_1,x_2,\ldots,d)$ has a root. By the above proposition, $f$ has a root iff one of these polynomials with fewer variables has a root.
Summary: We use real inequalities if $f$ has a term with all the variables, and divisibility otherwise, and that may be enough.