My answer is in addition to José's and Antinous's answers but maybe somewhat more abstract. In principle, their answers are using coordinates, whereas I'm trying to do it coordinate-free.

What you are looking for is the wedge or exterior product.
The exterior power $\bigwedge^k(V)$ of some vector space $V$ is the quotient of the tensor product $\bigotimes^k(V)$ by the relation $v\otimes v$.
To be somewhat more concrete and less abstract, this just means that for any vector $v\in V$ the wedge product $v\wedge v=0\in\bigwedge^2(V)$. Whenever you wedge vectors together, the result equals zero if at least two of the factors are linearly dependent.
Think of what happens to the cross product in $\mathbb{R}^3$.

In fact, let $e_1,e_2,\ldots,e_n$ be a basis of an inner product space $V$. Then $e_{i_1}\wedge e_{i_2}\wedge \ldots \wedge e_{i_k}$ is a basis for $\bigwedge^k(V)$ where $1\leq i_1 < i_2 < \ldots < i_k\leq n$.

If $V=\mathbb{R}^3$ then $v \wedge w$ equals $v \times w$ up to signs of the entries. This seems a bit obscure because technically $v\wedge w$ should be an element of $\bigwedge^2(\mathbb{R}^3)$. However, the latter vector space is isomorphic to $\mathbb{R}^3$. In fact, this relation is true for all exterior powers given an orientation on the vector space.
The isomorphism is called the Hodge star operator.
It says that there is an isomorphism $\star\colon\bigwedge^{n-k}(V)\to\bigwedge^{k}(V)$. This map operates on a $(n-k)$-wedge $\beta$ via the relation
$$
\alpha \wedge \beta = \langle \alpha,\star\beta \rangle \,\omega
$$
where $\alpha\in\bigwedge^{k}(V)$, $\omega\in\bigwedge^n(V)$ is an orientation form on $V$ and $\langle \cdot,\cdot \rangle$ is the induced inner product on $\bigwedge^{k}(V)$ (see wiki). Notice that the wiki-page defines the relation the other way around.

How does all this answer your question you ask?
Well, let us take $k=1$ and $V=\mathbb{R}^n$. Then the Hodge star isomorphism identifies the spaces $\bigwedge^{n-1}(\mathbb{R}^n)$ and $\bigwedge^{1}(\mathbb{R}^n)=\mathbb{R}^n$. This is good because you originally wanted to say something about orthogonality between a set of $n-1$ linearly indepedent vectors $v_1,v_2,\ldots,v_{n-1}$ and their "cross product".
Now let us exactly do that and set $\beta :=v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1}\in\bigwedge^{n-1}(\mathbb{R}^n)$. Then the image $\star\beta = \star(v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1})$ is a regular vector in $\mathbb{R}^n$ and the defining condition above implies for $\alpha=v_i\in\mathbb{R}^n=\bigwedge^{1}(\mathbb{R}^n)$
$$
v_i \wedge (v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1}) = \alpha \wedge \beta = \langle \alpha,\star\beta \rangle \,\omega = \langle v_i,\star\beta \rangle \,\omega.
$$
However, the left hand side equals zero for $i=1,2,\ldots,n-1$, so that the vector $\star\beta$ is orthogonal to all vectors $v_1,v_2,\ldots,v_{n-1}$ which is what you asked for. So you might want to define the cross product of $n-1$ vectors as $v_1 \times v_2 \times \ldots \times v_{n-1} := \star(v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1})$.

Maybe keep in mind that the other two answers implicitly use the Hodge star operation (and also a basis) to compute the "cross product in higher dimension" through the formal determinant which is encoded in the use of the wedge product here.