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.
First I'd recomend that you use more careful language in describing the groups and representations, including the precise notion of rank. The subject is complicated enough as is, so for instance you need to refer to the Lie algebras of special orthogonal groups as having Lie type $B_n$ or $D_n$ in the respective cases when $n$ is odd, even. (Then $n$ is the rank, while the natural representation of the group has dimension $2n+1$ or $2n$.)
Your question is actually extremely difficult to answer in detail, as people understood in the era around 1900 when invariant theory was being intensively developed. Eventually, in work of Hilbert, Weyl, and others, an emphasis was developed on ring-theoretic methods here. Rather than just concentrate on the multiplicity of the trivial module in each symmetric power of a given representation, one should look at the entire graded algebra of polynomials (or polynomial functions) and try to describe the (graded) subalgebra of invariants. The starting point is the "natural" representation of the Lie algebra or Lie group, where it's already quite challenging to make the answers explicit.
Modern sources include the Springer GTM volumes 129 by Fulton-Harris and 255 by Goodman-Wallach (a second edition of an earlier book published elsewhere). Relevant work by Howe and others is covered here, along with helpful examples and references.
Best Answer
The trivial representation appears in $\wedge^2 V$ if and only if the representation $V^{\ast}$ has a $G$-invariant alternating bilinear form (because $\wedge^2 V\cong\wedge^2\left(\left(V^{\ast}\right)^{\ast}\right)$ is isomorphic to the $G$-module of all alternating bilinear forms on $V^{\ast}$, and $G$-invariant forms correspond to $G$-fixed elements).
ctrl+c & ctrl+v:
The trivial representation appears in $\mathrm{Sym}^2 V$ if and only if the representation $V^{\ast}$ has a $G$-invariant symmetric bilinear form (because $\mathrm{Sym}^2 V\cong\mathrm{Sym}^2\left(\left(V^{\ast}\right)^{\ast}\right)$ is isomorphic to the $G$-module of all symmetric bilinear forms on $V^{\ast}$, and $G$-invariant forms correspond to $G$-fixed elements).
So we have to prove that for an irreducible representation $V$, the representation $V^{\ast}$ cannot have both a nontrivial $G$-invariant symmetric bilinear form and a nontrivial $G$-invariant alternating bilinear form. More generally, an irreducible representation $W$ of $G$ cannot have two linearly independent $G$-invariant bilinear forms. In fact, a bilinear form on the $k$-vector space $W$ can be seen as a homomorphism $W\to W^{\ast}$, and the bilinear form is $G$-invariant if and only if this homomorphism is $G$-equivariant. But Schur's lemma yields that there cannot be two linearly independent $G$-equivariant homomorphisms from $W$ to $W^{\ast}$, since both $W$ and $W^{\ast}$ are irreducible representations.
So much for the case when the ground field is algebraically closed (which is probably your case). In the general case, I think the assertion is not true, though I don't know a counterexample right out of my head.