The title and question have been edited in light of Ian Agol's comment. The previous question was stated in terms of the wrong notion of length to discuss deformations:
What is the longest slope $\gamma$ in the complement of Dehn surgery space of a cusped hyperbolic 3-manifold $M$? Here Dehn surgery space is the space of fillings such that the hyperbolic structure on the filling $M(\gamma)$ can be realized as a deformation of the original $M$.
To keep things focused, consider only fillings of one cusped manifolds.
This question is related to Ken Baker’s question:
Hyperbolic exceptional fillings of cusped hyperbolic 3-manifolds
However, Ken’s question is interested in the total number of slopes in this complement. This question is focusing on the longest such slope measured using normalized length, where the normalized length is measured by displacement in boundary of a horoball packing (as$\gamma$, $\mathcal{L}(\gamma)$ is as defined in the set up for the:
Hodgson, Craig D.; Kerckhoff, Steven P., $6$-theorem orUniversal bounds for hyperbolic Dehn surgery, Ann. Math. $2\pi$(2) 162, No. 1, 367-Theorem421 (2005). ZBL1087.57011.
Namely $\mathcal{L}(\gamma)=length(\gamma)/\sqrt{Area(\partial T)}$. Here, $length(\gamma)$ the translation length of $\gamma$ in a cusp neighborhood and $Area(\partial T)$ is the area of that torus in cusp neighborhood.
Of course, it is possible that this question as stated does not have a realizable answer, because there is no longest slope.
Here is a more carefully stated version:
What is the largest $L$$\mathcal{L}_{max}$ such that there exist a family of slopes $\gamma_i$ in (1-cusped hyperbolic) manifolds $M_i$ such that $$\lim_{i \to \infty} length(\gamma_i) = L$$$$\lim_{i \to \infty} \mathcal{L}(\gamma_i) = \mathcal{L}_{max}$$ and each $M_i(\gamma_i)$ is a hyperbolic manifold such that the hyperbolic structure cannot be realized as a deformation of the hyperbolic structure of $M_i$?
Of courseHodgson and Kerckhoff give an upper bound of $\mathcal{L}_{max}\leq C\approx 7.515$.
To give context how normalized length affects length, the fact that(3,3,3) pretzel knot has slope of length 6 yielding a torus filling. However, the normalized length of this slope is $\mathcal{L}=\frac{6}{\sqrt{A}}\approx 1.91673$, $A=\frac{8\sqrt{3}}{(1+3\sqrt{57})^{1/3}}+\sqrt{3}(1+3\sqrt{57})^{1/3}$. This example appears in
Adams, Colin; Bennett, Hanna; Davis, Christopher; Jennings, Michael; Kloke, Jennifer; Perry, Nicholas; Schoenfeld, Eric, Totally geodesic Seifert surfaces in hyperbolic knot and link complements. II, J. Differ. Geom. 79, No. 1, 1-23 (2008). ZBL1158.57004.
and as a filling of the manifold defined by Figure 6:
Agol, Ian, Bounds on exceptional Dehn filling, Geom. Topol. 4, 431-theorem is sharp implies that449 $L\geq 6$(2000). AlsoZBL0959.57009.
However, the cusp area appears to grow for $2\pi$ Theorem says(n,n,n)-pretzels, so I chose the $L\leq 2\pi$(3,3,3)-pretzel to seemingly get the longest slope (in terms of normalized length) in this family.