Let us take a relative category $(\mathcal{C},\mathcal{W})$, and consider its hammock localization $L\_H \mathcal{C}$. It seems to me that for every two objects $X,Y \in \mathcal{C}$ the mapping simplicial set $L\_H \mathcal{C}(X,Y)$ has the (strict) right lifting property against all inclusions $\partial \Delta^n \hookrightarrow \Delta^n$ for $n \geq 3$. The reason is the following: let us consider the case $n=3$ for simplicity, the general case is analogous. A diagram of shape $\partial \Delta^n$ means a collection of $n+1$ hammocks of width $n-1$ such that some selected $(n-2)$-faces of these correspond. Therefore, up to unreducing these hammocks, they can all be assumed to be of the same length, so let us consider only one column at a time. For each $i=0,1,2,3$ we have four columns of the form


$$\require{AMScd}
\begin{CD}
\bullet @>f\_1^0>> \bullet\\
@Vv\_{12}^0VV @VVw\_{12}^0V\\
\bullet @>f\_2^0>> \bullet\\
@Vv\_{23}^0VV @VVw\_{23}^0V\\
\bullet @>f\_3^0>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0^1>> \bullet\\
@Vv\_{02}^1VV @VVw\_{02}^1V\\
\bullet @>f\_2^1>> \bullet\\
@Vv\_{23}^1VV @VVw\_{23}^1V\\
\bullet @>f\_3^1>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0^2>> \bullet\\
@Vv\_{01}^2VV @VVw\_{01}^2V\\
\bullet @>f\_1^2>> \bullet\\
@Vv\_{13}^2VV @VVw\_{13}^2V\\
\bullet @>f\_3^2>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0^3>> \bullet\\
@Vv\_{01}^3VV @VVw\_{01}^3V\\
\bullet @>f\_1^3>> \bullet\\
@Vv\_{12}^3VV @VVw\_{12}^3V\\
\bullet @>f\_2^3>> \bullet
\end{CD}
$$


I'm omitting names of objects, a superscript index $i$ just refers to the $i$-th labeled 2-face, a subscript index $i$ refers to the $i$-th row, and a subscript index $ij$ refers to the arrow going from the $i$-th row to the $j$-row. Now let us look at what it means for these wannabe $2$-faces to be assembled into a diagram of shape $\partial \Delta^3$.  

For instance, the 0-th face of the 0-th diagram and the 0-th face of the 1st diagram should coincide, and similarly for the other compatibility condtions. This allows us to remove the superscript indices from all $f$'s and from all $v\_{ij}$'s and $w\_{ij}$'s whenever $i$ and $j$ are consecutive.  

Moreover, the 1st face of the 0-th diagram and the 0-th face of the 2nd diagram should coincide, and similarly for others. This means that $v\_{13}^2 = v\_{23} \circ v\_{12}$ and analogously for $w\_{13}^2$. This means that we can also remove the subscript indices $ij$ when they are not consecutive and write the corresponding $v$'s and $w$'s in terms of those with consecutive indices. In other words, the above columns may be rewritten as


$$\require{AMScd}
\begin{CD}
\bullet @>f\_1>> \bullet\\
@Vv\_{12}VV @VVw\_{12}V\\
\bullet @>f\_2>> \bullet\\
@Vv\_{23}VV @VVw\_{23}V\\
\bullet @>f\_3>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{12}\circ v\_{01}VV @VVw\_{12} \circ w\_{01}V\\
\bullet @>f\_2>> \bullet\\
@Vv\_{23}VV @VVw\_{23}V\\
\bullet @>f\_3>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{01}VV @VVw\_{01}V\\
\bullet @>f\_1>> \bullet\\
@Vv\_{23} \circ v\_{12}VV @VVw\_{23} \circ w\_{12}V\\
\bullet @>f\_3>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{01}VV @VVw\_{01}V\\
\bullet @>f\_1>> \bullet\\
@Vv\_{12}VV @VVw\_{12}V\\
\bullet @>f\_2>> \bullet
\end{CD}
$$


Now consider the column


$$\require{AMScd}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{01}VV @VVw\_{01}V\\
\bullet @>f\_1>> \bullet\\
@Vv\_{12}VV @VVw\_{12}V\\
\bullet @>f\_2>> \bullet\\
@Vv\_{23}VV @VVw\_{23}V\\
\bullet @>f\_3>> \bullet
\end{CD}
$$


whose faces are clearly the four columns above. Reasoning like this for each column, we have constructed a diagram of shape $\Delta^3$ whose restriction to $\partial \Delta^3$ is what we started with.


There should be a mistake in the above argument, in that we know that there are relative categories with arbitrary mapping simplicial sets, but I fail to find it. Could someone please point it out?



Are hammock localizations locally truncated?
Your calculation is correct. For every two objects $X, Y \in \mathcal{C}$, the hom space $L^H\mathcal{C}(X,Y)$ has the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$.


First note that the nerve of any category has the strict right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$. The space $L^H\mathcal{C}(X,Y)$ is not the nerve of a category, but it is a quotient of the nerve a category (See [here](https://ncatlab.org/nlab/show/simplicial+localization#DwyerKanCalculating)). Call this nerve $N\mathcal{D}(X,Y)$. Your argument can be summarized by saying that any map $\partial \Delta^n \to L^H\mathcal{C}(X,Y)$ can be factored through $N\mathcal{D}(X,Y)$, where we can solve the lifting problem and then project back into $L^H\mathcal{C}(X,Y)$. (This is the "unreducing" part of your argument).


Note that what I just said does not yet prove that the lift to $L^H\mathcal{C}(X,Y)$ is unique. This would require a further argument about what happens when you have two different lifts $\partial \Delta^n \to N\mathcal{D}(X,Y)$. This might be possible - I am not 100% sure though.


In any case, the simplicial set $L^H\mathcal{C}(X,Y)$ is not a Kan complex (usually). So, as pointed out in the comments by Zhen Lin, having the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$ does not tell us about the higher homotopy groups of $L^H\mathcal{C}(X,Y)$. Indeed, as you correctly observe, there are relative categories which allow you to realize $L^H\mathcal{C}(X,Y)$ as any homotopy type you like.


