I'll share what I learned after reading many papers about the subset sum problem, I too implemented a solver and was looking time ago to understand what a 'hard instance' really is (in regards to this problem), and what I'm going to describe worked very well as a test bench for my solver, I personally never tackled the backtracking algorithm so you might need to reconcile this information with BT dynamics, but on the other hand, this way to build up instances are effective for both DP and the 'meet-in-the-middle' approach.
In relationship to subset sum problem, the existing literature, defines usually a hard instance as a random set of equally distributed values across the problem space with $density \approx 1$ (I found a paper settling, after some analysis that in fact the hardest instances have a $density \approx 1.03$ but is not a big of a difference form the practical point of view, that is what it looks you are interested)
Assuming $N$ being the number of elements and $P$ the number of bits for encoding the maximum possible value in your set, to achieve a $density \approx 1$ you must set the same value for $N$ and $P$ as $density=N/P$
For example, as you are constraining the integers resolution to no more than 64 bits, then, $N=P=64$.
Now you need to pick 64 uniquely random values equally distributed between $1..2^{64}$, trying to really equally distribute them (not clustering near-values). This will be your input set.
Then, you need to pick the target value, for this, you must select, again, randomly, unique values from your set, and add them until the sum is around the 50% of the problem space (problem space=the sum of all your values).
Why around 50%? Because if you pick a target value over 50% of the problem space, let's say 80%, by using DP, you can just build up the sums for the subsets and check for the 'negative' subset, as probably you can do with BT, so any target value deviating from the middle location is not that harder (because is near the beginning or the end, in the middle is equidistant, thus is the hardest of the locations for the target value)
However, in my opinion, the concept of 'hard instance' for a subset sum solver depends a lot on the solver implementation details, each solver can face problems with different parameters, e.g: DP can't handle a big $P$ because it blows up the memory need to hold the dynamic table in memory, even with very smart optimizations, the DP 'cost' depends on the target value, and in its worst case the target value for DP is a value around $2^{P-1}*N$, so any change on $P$ or $N$ will impact the 'cost'.
If we look to the meet-in-the-middle, $P$ is not that important (if we don't consider memory), and $N$ is the real problem, when you get to larger sets, even with the smart split the 'cost' will increase exponentially.
So the bottom line is to know what variables impact your algorithm performance to be able to create proper hard instances for 'test benching'