This was meant to be a comment, but it won't fit, so here we go.
Just a few naive observations.
For completeness:
Weak BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have:
$$\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$$
BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have:
(1) $\text{Sha}(A/K)$ is a finite abelian group.
(2) $\rho :=\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$
(3) $\lim_{s\to 1}L(A/K,s)(s-1)^{-\rho} = \frac{R_A\cdot\Omega_A\cdot\#\text{Sha}(A/K)}{|\Delta_K|^{1/2}\cdot \# A(K)_{\rm tor}\cdot\# A^{\vee}(K)_{\rm tor}}.$
Super-weak BSD The proportion of abelian varieties defined over $K$ that satisfy Weak BSD above, is 100%.
If $K$ is the global function field of a smooth projective geometrically irreducible curve over $\mathbf{F}_q$, then it is known by work of Tate, Milne, Bauer, Schneider, Kato, Trihan, that Weak BSD is equivalent to BSD, and BSD is in turn equivalent to finiteness of the $\ell$-primary torsion in the Tate-Shafarevich group for some prime $\ell$ (allowed to be the characteristic of $K$).
If $K$ is a number field, the above stream of equivalences is still expected to be true, but not known yet, likely because there's still no robust cohomological method to put us in the position to mimic the flat cohomology/crystalline-syntomic cohomology methods employed in the positive characteristic case, which Iwasawa theory for abelian varieties was partially meant for. Let us grant for a moment the following:
Expectation Let $K$ be a number field. For any abelian variety $A$ defined over $K$, the following are equivalent:
(1) $\text{Sha}(A/K)$ is finite.
(2) For all primes $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite.
(3) For some prime $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite.
(4) Weak BSD is true for $A$.
(5) Strong BSD is true for $A$.
This expectation, a Theorem if $K$ were a positive characteristic global function field, reveals the problem is really about showing finiteness of $\text{Sha}(A/K)(\ell)$.
Over number fields, it is not even known that $\text{Sha}(A/K)(\ell)$ vanishes for almost all primes $\ell$.
The takeaway from the Bhargava-Skinner-Zhang paper, and from the answers and comments here, is that knowledge of finiteness $\text{Sha}(A/K)(\ell)$ does not actually help much or at all to improve their progress on Super-weak BSD ($\text{Sha}$-finiteness enters through the parity conjecture, a theorem unconditionally, and definitely a much weaker statement than finiteness of $\text{Sha}(A/K)(\ell)$), which to me just means such methods fail to get to the point of the problem itself, and will not solve it.
In other words, I don't see any trace of the ability to produce interesting algebraic cycles on abelian varieties, into them, and according to the Expectation above, true in char $p$ though open in char $0$ so far, this should be the whole point.
Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional.
Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.
EDIT: I should also add that several of the averages (both unconditional and conjectural) that are key to the Bhargava-Skinner-Zhang methods, fail in char $p$, while I'd regard a method towards BSD to be "promising", if it were able to make progress or settle its char $p$ counterpart, first.
Best Answer
No, there are no such examples known. In fact, with the current technology, the two questions are more or less equally hard. That's because for any given prime $p$, you can, in principle, establish finiteness of $Ш(E/\mathbb{Q})[p^\infty]$ algorithmically by performing $p^n$-descent for higher and higher $n$, until the upper bound on the rank of $Ш(E/\mathbb{Q})[p^n]$ stabilises. Of course, we cannot prove a priori that this would ever happen, but in practice, if you knew finiteness of $p$-primary parts of sha outside a finite set of primes, you would run your computer to do $p^n$-descent for the remaining primes, until you establish finiteness for this finite set, too.