For $p=2$, the plethysm is known. A reference is Macdonald's book "Symmetric Functions and Hall Polynomials" (p.138, example 6d in the second edition).
$\wedge^\bullet( Sym^2) = \bigoplus_\lambda \Sigma^\lambda$ where the sum is over all $\lambda$ with the following property: for each box on the main diagonal of its Young diagram, you have exactly one less box in the column directly below it (not counting the diagonal) than there are boxes in the row directly to the right of it (again, not counting the diagonal), and the sum is multiplicity-free. To get the case of $\wedge^q Sym^2$ just take those partitions of size $2q$ in the set above. So for example, the first few which are hooks are:
$0, (2), (3,1), (4,1,1), (5,1,1,1), (6,1,1,1,1), \dots$
but you can also get things which are unions of more hooks. Here are some examples with two hooks:
$(3,3), (4,3,1), (5,3,1,1), (6,3,1,1,1), \dots, (4,4,2)$
Another way to describe it: if you have a partition in the set I described, you are allowed to add a box both to the $i$th row and $i$th column of its Young diagram to get another partition in the set (assuming this creates a valid partition).
For a specific $T$, just take all $\lambda$ with at most $k$ parts (all other Schur functors vanish). The case $p>2$ is essentially a "wild" situation and you probably shouldn't expect any workable formulas except when $q$ or $k$ is small.