When $d=2$, the decomposition is known for all $k$ and $n$. Given a partition $\lambda$ of $k$ with distinct parts, let $2[\lambda]$ denote the partition of $2k$ whose main-diagonal hook lengths are $2\lambda_1, \ldots, 2\lambda_k$, and whose $i$th part has length $\lambda_i + 1$. Then

$$ \bigwedge^k {\rm Sym}^2 V = \sum_\lambda S^{2[\lambda]}(V) $$

where the sum is over all partitions $\lambda$ with distinct parts such that $2[\lambda]$ has at most $n$ parts and $S^\mu$ is the Schur functor for the partition $\mu$. For a proof using the symmetric group see Lemma 7 in http://arxiv.org/abs/0903.2864.

<b>Edit (June 2014).</b> The constituents of $\bigwedge^3 \mathrm{Sym}^{d}(V)$ are determined on page 141 of Macdonald's book, Symmetric functions and Hall polynomials. Remark 3.6(b) in Howe, <em>$(GL_n,GL_m)$-duality and symmetric plethysm</em>, Proc. Indian Acad. Sci. <b>97</b> (1987) 85–109, gives a method for computing the plethysm $\bigwedge^4 \mathrm{Sym}^d(V)$. 

Apart from the case $k=2$ mentioned in this question (and the trivial cases $k=1$ or $d=1$), I think these are the only case where the complete decomposition is known.