**Edit after a fruitful [discussion][1] with Marco Golla:** 

1. My claim "$\Sigma(2,3,19) = S^3_{+1}(8_1)$" is based on Example 1.4 of Saveliev's book [Invariants of Homology 3-Spheres][2].

2. According to [Theorem 1.1. (3)][3] of Brittenham and Wu, $S^3_{+1}(8_1)$ is hyperbolic and $S^3_{-1}(8_1)$ is Seifert fibered.

It seems that there is a sign/orientation inconsistency somewhere...

3. Therefore, the "Theorem" at the end of (c) part of my answer is still a conjecture for $n \geq 5$. The rest will be useful in the future, so I keep the answer same.

o____________________________________________________________________________

Golla and Hayden gave awesome responses to the question. Their arguments can be generalized in the following fashion.

Following their notations, recall that $K_1$ is the figure-eight $4_1$ knot and in general $K_n$ is the twist knot $(2n+2)_1$ in $S^3$. 

They together proved $$S^3_{1/3}(K_1) \cong S^3_{+1}(K_3).$$

Note that the right-hand side is the Brieskorn sphere $\Sigma(2,3,19)$.

a- Handle diagrams of Golla is generalized to the next case as follows:

[![enter image description here][4]][4]

b- With the observations of Hayden, we have $$S^3_{1/4}(K_1) \cong S^3_{+1}(K_4).$$ There is a pattern in the Kirby calculus diagrams. Thus we may eventually prove that $$S^3_{1/n}(K_1) \cong S^3_{+1}(K_n).$$ Similarly, the right-hand side is the Brieskorn sphere $\Sigma(2,3,6n+1)$.

c- This part is about rational homology cobordism classes of $\Sigma(2,3,6n+1)$.

**Definition:** A knot $K$ in $S^3$ is called *rationally slice* if it bounds a smoothly properly embedded disk $D$ in a rational homology ball $X$.

**Theorem(Kawauchi, [(Kaw79)][5] + [(Kaw09)][6])** Any hyperbolic amphichiral knot in $S^3$ is rationally slice. Consequently, $K_1$ is rationally slice in $S^3$.

Now we need an extra observation which is probably known to experts in low-dimensional topology and can be seen as the rational analogue of Gordon's theorem:

**Lemma:** For each $n$, $S^3_{1/n}(K_1)$ bounds a rational homology ball.

**Proof:** The figure-eight knot $K_1$ bounds a smooth disk $D$ in a rational homology ball $X$. The tubular neighborhood of $D$, $\nu(D)$, is $B^2 \times D$ in $X$. 

Think $K_1$ and $D$ respectively as a belt sphere and co-core of $4$-dimensional $2$-handle $B^2 \times B^2$. So, we have $B^2 \times D = (X \setminus \nu(D))⋃  B^2 \times B^2.$ 

Now remove this $2$-handle and reattach it with a framing differing from the initial one by $n$ left-handed twists. Then the boundary $3$-manifold changes by $1/n$-surgery on $K_1$. Since we don't change the rational homology of $4$-manifold, we are done.

Therefore, we have a "theorem":

**Theorem:** For each $n$, Brieskorn spheres $\Sigma(2,3,6n+1)$ bounds a rational homology ball.

**Remark:** The cases $n=1$ and $n=3$ are known by Fintushel-Stern [(FS84)][7] and Akbulut-Larson [(AL18)][8]. For the cases $n=2$ and $n=4$, they bound contractible $4$-manifolds due to classical results of Akbulut-Kirby [(AK79)][9] and Fickle [(F84)][10]. Hence they a priori bound rational homology balls.


  [1]: https://chat.stackexchange.com/rooms/110388/discussion-between-marco-golla-and-oguz-savk
  [2]: https://www.springer.com/gp/book/9783540437963
  [3]: https://arxiv.org/pdf/math/0011005.pdf
  [4]: https://i.sstatic.net/NP0fx.png
  [5]: https://projecteuclid.org/download/pdf_1/euclid.pja/1195517071
  [6]: https://pdfs.semanticscholar.org/e454/67e607a0eafef7be8dd6f6262879c06ef5a9.pdf?_ga=2.177894508.83553317.1594207656-1828353817.1594207656
  [7]: http://faculty.sites.uci.edu/rstern/files/2011/03/20_mu_invariant_one_sphere.pdf
  [8]: https://arxiv.org/pdf/1704.07739.pdf
  [9]: https://projecteuclid.org/download/pdf_1/euclid.mmj/1029002261
  [10]: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.449.3854&rep=rep1&type=pdf