For $SO(n)$ a calculation using LiE gives:
(using partition notation so $W$ is [2])
and assuming $n$ is not small
For $W\otimes W$, [4],[3,1],[2,2],[2],[1,1],[]
(all with multiplicity one)
and for $W\otimes W\otimes W$,
1.[6] 2.[5,1] 3.[4,2] 1.[3,3] 1.[4,1,1] 2.[3,2,1] 1.[2,2,2] 3.[4] 6.[3,1] 2.[2,2] 3.[2,1,1] 6.[2] 3.[1,1] 1.[]
The same works for $Sp(n)$ by taking conjugate partitions.
There is also a relationship with $SL(n)$.
This is taking your question at face value. If it is understanding you're after instead then the best approach is to use crystal graphs.
The notation I have used denotes a representation by a partition. I have put $m.$ in front to denote multiplicity is $m$. A partition is $[a_1,a_2,a_3,...]$ where $a_i\ge a_j$ if $i$ less than $j$. To convert to a highest weight vector add the appropriate number of $0$s to the end.
Then take $[a_1-a_2,a_2-a_3,a_3-a_4,...]$. This gives a dominant integral weight. The fundamental weights are the partitions $[1,,,,1]$. If this has length $k$ this corresponds to the $k$-th exterior power of the vector representation (provided $2k-1$ less than $n$).
In particular the trivial representation is $[]$, the vector representation $V$ is $[1]$, the exterior square of $V$ is $[1,1]$, the symmetric square is $[2]+[]$.
For the $k$-th tensor power of $W$ you will see partitions of $2k-2p$ for $0\le p\le k$ only and it remains to determine the multiplicities (possibly $0$). For $SL(n)$ just take the partitions of $2k$ (with their multiplicities) and ignore the rest.
If I remember this correctly the cases $\mathrm{Sym}^k(\bigwedge^2 \mathbb{C}^n)$ and $\mathrm{Sym}^k(\mathrm{Sym}^2(\mathbb{C}^n))$ are known; and hence $\bigwedge^k(\bigwedge^2 \mathbb{C}^n)$ and $\bigwedge^k(\mathrm{Sym}^2(\mathbb{C}^n))$. I will look up the references tomorrow (if this is of interest).
Edit The result has now been stated. I learnt this from R.P.Stanley "Enumerative Combinatorics" Vol 2, Appendix 2. Specifically, A2.9 Example (page 449) which refers
to (7.202) on page 503. This gives as the original reference (11.9;4) of the 1950 edition of:
Littlewood, Dudley E.
"The theory of group characters and matrix representations of groups."
P.S. In the Notes at the end of 7.24 (bottom of page 404 in CUP 1999 edition)
it discusses the origin and the etymology of "plethysm". It says:
Plethysm was introduced in
MR0010594 (6,41c) Littlewood, D. E. Invariant theory, tensors and group characters.
Philos. Trans. Roy. Soc. London. Ser. A. 239, (1944). 305--365
The term "plethysm" was suggested to Littlewood by M. L. Clark after the Greek word
plethysmos $\pi\lambda\eta\theta\upsilon\sigma\mu\acute o\varsigma$ for "multiplication".
Best Answer
Have a look at the beginning of section 33 (in particular, 33.2) of the book "Natural operations in differential geometry" (pdf), for the Riemannian case only. It should work for the $SO(p,q)$ case also. There all $O(n)$-invariant tensors are described: The idea is to tensor with the metric or its inverse and then use the $GL(n)$ decomposition, i.e., involve traces and permutations.
Edit: Link corrected.