As others have pointed out the main problem is what "taking a vector at random" means. Probability theory requires that one specifies a certain probability measure on ${\mathbb R}^n$ before one can make any predictions about outcomes of experiments concerning chosen vectors. E.g., if it is totally unlikely, meaning: the probability is zero, that a vector with $x_n\ne 0$ is chosen, then the probability that $n$ vectors chosen independently are linearly independent is $\>=0$, since with probability $1$ they all lie in the plane $x_n=0$.

A reasonable starting point could be installing a *rotational invariant* probability measure. As the length of the $n$ chosen vectors does not affect their linear dependence or independence this means that we are chosing $n$ independent vectors uniformly distributed on the sphere $S^{n-1}$. (This informal description has a precise mathematical meaning.)

Under this hypothesis the probability that the $n$ chosen vectors $X_k$ are linearly independent is $=1$.

*Proof.* The first vector $X_1$ is linearly independent with probability $1$, as $|X_1|=1$. Assume that $1< r\leq n$ and that the first $r-1$ vectors are linearly independent with probability $1$. Then with probability $1$ these $r-1$ vectors span a subspace $V$ of dimension $r-1$, which intersects $S^{n-1}$ in an $(r-2)$-dimensional "subsphere" $S_V^{r-2}$. This subsphere has $(n-1)$-dimensional measure $0$ on $S^{n-1}$. Therefore the probability that $X_r$ lies in this subsphere is zero. It follows that with probability $1$ the vectors $X_1$, $\ldots$, $X_{r-1}$, $X_r$ are linearly independent.