Here's at least a couple of candidate designs which have the first four properties, although I haven't checked acyclicity and don't yet see how to exploit the existing symmetries to check it other than by a brute-force rank computation of the boundary map.

(1) The orbit of the set $\{ \infty, 0, 1, 2 \}$ under the fractional-linear action of $G=\mathrm{PSL}_2(11)$ on $\mathbb{Z}/11\mathbb{Z} \cup \{ \infty \}$ gives a $3$-$(12,4,3)$ design, which we can decorate with orientations to pass from subsets to simplices to chains. The $15$ blocks containing the $1$-face $\{ \infty, 0 \}$ induce a bipartite $3$-regular graph of order $10$ on the complement of this face; each edge of this graph connects a quadratic residue modulo $11$ to a non-residue (but not all such pairings occur); multiplication by the quadratic residue $3$ (which is in $G$ as $\mathrm{diag}(6,2)$) induces a symmetry of order $5$ on this graph. The $2$-face $F=\{ \infty, 0, 1 \}$ extends to the blocks $\{\infty 0 1 2\}$, $\{\infty 0 1 6\}$, $\{\infty 0 1 a\}$ (writing $a$ for the digit "10"), and the stabilizer in $G$ of $F$ as a set, generated by $x\mapsto 1-1/x$, permutes these blocks. And $G$ acts transitively on the $220$ three-element subsets, as well as on the two-element ones, so everything is as symmetric as one can hope for.

(2) Tung-Hsin Ku in *Simple BIB Designs and 3-Designs of Small Orders* (in W.D.Wallis *et al* (ed.), *Combinatorial Designs and Applications*, Lect. Notes in Pure and Applied Math. **126** (1990), 79-85) describes among others a $3$-$(12,4,3)$ design on the same set on which only $\mathbb{Z}/11\mathbb{Z}$ acts, leaving $\infty$ fixed. The blocks are $\{0 1 2 3\}$, $\{0 1 2 4\}$, $\{0 1 3 7\}$, $\{0 1 4 5\}$, $\{0 1 4 8\}$, $\{0 1 5 6 \}$, $\{0 1 6 9\}$, $\{0 2 4 6\}$, $\{0 2 5 7\}$, $\{0 2 5 8\}$, $\{0 1 5 \infty\}$, $\{0 1 8 \infty\}$, $\{0 1 9 \infty\}$, $\{0 2 5 \infty\}$, $\{0 2 7 \infty\}$ and their images under the group action. Here, proceeding as above, several different $3$-regular graphs of order $10$ occur among the complements of $1$-faces. Some of these graphs contain one or more triangles, but in no case are all five blocks present that would combine to the $3$-skeleton of a single $4$-simplex. (Which doesn't rule out more complicated cycles.)

**Added 2016-07-30:**

Computing the rank and the kernel of the boundary map in the two examples by brute force is unexciting but straightforward, so (for lack of a better idea and to see what would happen) I went through the exercise. It helps to organize the matrix as a 20-by-15 array of 11-by-11 circulants, most of which are zero and all of which are sparse.

If I didn't get any signs wrong, (1) indeed yields ~~an acyclic complex~~ (*Edit 2016-08-02*) a complex ~~with $H_3(K,\mathbb{Z})=0$, $H_2(K,\mathbb{Q})=0$, and finite $H_2(K,\mathbb{Z})$~~. (**Edit 2017-10-18**) It seems I *did* get a sign wrong in one of the 11-by-11 circulants (and the first extra consistency check I thought of today but hadn't thought of last year immediately found it). Rerunning the rank computation, this example is now far from acyclic: the interesting boundary map has rank 155 instead of 165, thus both $H_3(K,\mathbb{Q})$ and $H_2(K,\mathbb{Q})$ will be 10-dimensional.

Ku's design (2) doesn't result in an acyclic complex; here I got $H_3(K,\mathbb{Z}) \cong \mathbb{Z}^2$. Generating cycles are sums over entire $\mathbb{Z}/11\mathbb{Z}$ orbits (and not involving the fixed point of the group action):
$$\begin{gather}
\{0145\}+\{1256\}+\dotsb-\{0148\}-\{1259\}-\dotsb+\{0156\}+\{1267\}+\dotsb; \\
\{0246\}+\{1357\}+\dotsb+\{0257\}+\{1368\}+\dotsb-\{0258\}-\{1369\}-\dotsb.
\end{gather}$$
(*Edit 2017-10-18*) There might be a sign error here, too, affecting the outcome - there might be yet more cycles apart from the above. I don't have time to re-check this one right now.

Thus with a third of the 3-simplices taking part in the game, we seem to be near the border of where there are enough of them to combine into nontrivial cycles. I wouldn't be surprised if there were other solutions with fewer symmetries or with none at all.

(For comparison, the analogue of (1) using $\mathrm{PSL}_2(7)$ and the orbit of $\{\infty 012\}$ results in a $3$-$(8,4,3)$ design with $42$ blocks among the seventy $3$-simplices and an $H_3(K,\mathbb{Z})$ of rank $8$, spanned by the $24$-element orbit of
$$\{0123\}+\{1234\}+\dotsb+\{6012\}+\{0134\}+\{1245\}+\dotsb+\{6023\},$$
which is again a sum over a whole orbit of the affine $\mathbb{Z}/7\mathbb{Z}$. Here we seem to have too many blocks to avoid such combinatorial coincidences. [Note that this is *not* the case $k=2$ of the question.])

A candidate for the case $k=4$, thus $n=20$, might be constructed in terms of $\mathrm{PSL}_2(19)$, remembering that the action by fractional linear transformations will be far from $4$-transitive; thus multiple orbits must be used to arrive at the desired $3876$ blocks. For $k=5$ and $k=6$, we also have convenient primes $29$ and $41$ to play this kind of game, moving even further beyond the available symmetries. But $55$ isn't prime, so a new idea would be needed for $k=7$.

*Added 2017-10-17:*
By popular request, the 165 blocks for the first example arise from letting the affine $\mathbb{Z}/11\mathbb{Z}$ act on the 15 representatives $\{\infty 0 1 2\}$, $\{\infty 0 1 6\}$, $\{\infty 0 2 4\}$, $\{\infty 0 3 6\}$, $\{\infty 0 3 7\}$, $\{0 1 2 5\}$, $\{0 1 2 8\}$, $\{0 1 3 5\}$, $\{0 1 3 7\}$, $\{0 1 3 8\}$, $\{0 1 4 6\}$, $\{0 1 4 9\}$, $\{0 1 5 9\}$, $\{0 1 6 8\}$, $\{0 1 7 9\}$.