Let $A$ be the matrix whose columns are the vectors $v_1, v_2, ... v_n$. Then the Gram matrix is $A^T A$, so $\det G = (\det A)^2$.
Edit: Here's an explanation that ignores the dimension of the ambient space. If $\langle \cdot, \cdot \rangle$ denotes the inner product, then the Gram matrix is precisely the matrix describing the inner product
$\langle x_1 v_1 + ... + x_n v_n, y_1 v_1 + ... + y_n v_n \rangle$
on $\mathbb{R}^n$. It's not hard to see that the vectors $v_i$ are linearly independent if and only if the above inner product is positive-definite. But we can write the above as $v^T Gv$ where $v \in \mathbb{R}^n$ and $G$ is the Gram matrix, and then we know that the inner product is positive-definite if and only if $G$ is invertible (since $G$ is invertible if and only if its eigenvalues are all positive).