$$$\dfrac{1}{2^k}$$$

Let $$$n$$$ be the length of $$$A$$$. Numbers from $$$0$$$ to $$$2^k-1$$$ with $$$xor$$$ is a $$$k$$$-dimensional vector space over $$$\mathbb{F}_2$$$, therefore if $$$n \gt k$$$ it is guaranteed that $$$A$$$ has $$$0$$$-sum subsequence, because $$$A$$$ is linearly dependent collection of vectors.

If $$$n\leq k$$$ we can calculate the probability by counting number of linearly independent collections of length $$$n$$$: there are $$$2^k - 1$$$ ways to choose the fisrt vector, $$$2^k-2$$$ ways for the second... $$$2^k-2^t$$$ for the $$$t+1$$$'th vector, so overall number is $$$\prod_{i=1}^{n}(2^k-2^{i-1})$$$.

Then the needed probability is $$$1- \frac{\prod_{i=1}^{n}(2^k-2^{i-1})}{2^{nk}}$$$.

Correct me if I am wrong.