As far as I am aware, nothing unconditional is known. The difficulty is that one has no obvious constraint on the size of the variables, so that the $x_i$ could be arbitrarily large in terms of $n$ in a solution. The difficulty of ruling out solutions (when congruence conditions do not rule out solubility) is related to proving insolubility of generalised Fermat equations.

However, there is a conditional approach assuming the truth of the generalised ABC Conjecture (or, as Pomerance describes it, the Alphabet Conjecture). Let's stick with the general situation with $v$ summands. Generalised ABC asserts that in any solution of
$a_1+\ldots +a_s=0$ in which there are no vanishing subsums, one has
$\underset{1\le i\le s}{\max}|a_i|\ll_{s,\epsilon} \left( \prod_{p|a_1\ldots a_s}p\right)^{(s-1)(s-2)/2+\epsilon}.$ There is disagreement about the specific exponent, but this does not matter so much in the conclusion (see a 1986 paper of Brownawell and Masser for a function field version, and I discuss this in a 1994 paper on Quasi-diagonal behaviour). The vanishing subsums in your equation $\pm x_1^k+\ldots +\pm x_v^k=n$ make life easier (treat them separately, and more powerful estimates are possible), so for simplicity suppose that the representations all have no vanishing subsums. Then one obtains $|x_i|^k\ll |nx_1\ldots x_v|^{v(v-1)/2+\epsilon}$. Provided that $k>v^2(v-1)/2$, it follows that $\max |x_i|\ll |n|^{\alpha+\epsilon}$, where $\alpha=v(v-1)/(2k-v^2(v-1))$. OK ... so far so good. What we have shown thus far is that the variables in a representation are bounded by $|n|^{\alpha +\epsilon}$. The total number of variables available to represent the integers $n$ between $N/2$ and $N$ is consequently no larger than a quantity which is $\ll (N^{\alpha+\epsilon})^v$ (there were $v$ variables). Whenever $v(\alpha+\epsilon)<1$, therefore, the set of integers represented must have zero density. If I have not made any computational errors along the way, this leads us to the conclusion that whenever $k>v^2(v-1)$, then the density of integers represented in the easier Waring problem will be zero. (But remember that this is all conditional on the Generalised ABC Conjecture.) For the specific problem with $v=5$, it looks as if $k>100$ conjecturally does the trick (though smaller $k$ should surely also work).