The idea is to think of each layer outside of the [core] as a rotatable disk and then only count a single member from each of the resulting equivalence classes, which I think can be done by requiring that the addends are non-increasing from the core to the surface as well as when proceeding counter-clockwise from $0$ rad. For example, the invalid partitions
11[2]2 2[2]11
of $6$ are equivalent to
1[2]21
For $1,2,3,4,5$ I count the following $1,2,4,8,15$ partitions (resp.):
[1]
[2] [1]1
[3] [2]1 [1]11 1[1]1
1
[4] [3]1 [2]2 [2]11 1[2]1 [1]111 1[1]11 1[1]1
1 1 1
[5] [4]1 [3]2 [3]11 1[3]1 [2]21 1[2]2 [2]111 1[2]11 1[2]1 [1]1111 1[1]111 11[1]11 1[1]11 1[1]1
1
I also counted $28$ partitions of $6$, but I'm not sure if that number is right. I haven't found any promising leads in oeis, but it seems likely that something is known about this kind of partition.
I'm hoping for further terms for the corresponding partition function, a generating function, or a source with more information.
Edit: I realized that the circular version of Young tableaux is essentially the same as the original. That is, if you flip the Young tableaux for a shape with $n$ cells upside-down and wrap it around $n+1$, you get the circular Young tableaux for the corresponding circular Young diagram. I don't think that says much about this problem, but perhaps there is a nice connection to plane partitions if we unwrap our partitions. In particular (if $n$ is the number we're partitioning), I guess we want plane partitions of $n-k-a$ for $a\geq0$ and with $k$ the greatest part of the partition.
As you started to say in your edit, this all comes down to plane partitions.
An expression for the number of your partitions of $n$ is $$ \sum_{c=1}^n pp(n-c,c) $$ ($c$ for [core]) where $pp(i,j)$ is the number of plane partitions of $i$ with largest part $j$; this triangle of numbers is given in OEIS A242642 (with references back to MacMahon), which is the cumulative version of A091355. Note that if $j\geq i$ then $pp(i,j) = pp(i)$, the (unrestricted) number of plane partitions of $i$. Here are the next few values for the number of your partitions. $$6: pp(5,1) + pp(4,2) + pp(3) + pp(2) + pp(1) + pp(0) = 7 + 10 + 6 + 3 + 1 + 1 = 28.$$ $$7: pp(6,1) + pp(5,2) + pp(4,3) + pp(3) + pp(2) + pp(1) + pp(0) \\ = 11 + 16 + 12 + 6 + 3 + 1 + 1 = 50.$$ $$8: pp(7,1) + pp(6,2) + pp(5,3) + pp(4) + pp(3) + pp(2) + pp(1) + pp(0) \\ = 15 + 29 + 21 + 13 + 6 + 3 + 1 + 1 = 89.$$ $$9: pp(8,1) + pp(7,2) + pp(6,3) + pp(5,4) + pp(4) + pp(3) + pp(2) + pp(1) + pp(0) \\ = 22 + 45 + 40 + 23 + 13 + 6 + 3 + 1 + 1 = 154.$$ $$10: pp(9,1) + pp(8,2) + pp(7,3) + pp(6,4) + pp(5) + pp(4) + pp(3) + pp(2) + pp(1) + pp(0) \\ = 30 + 75 + 67 + 45 + 24 + 13 + 6 + 3 + 1 + 1 = 265.$$
(The computation for $n=7$ in my comment was off by 1 for not counting 2 2 | 1 and 2 1 | 2 separately in the cases with [2].) Curiously, the enumeration of your partitions matches A036615 up to $n=8$ but that sequence's next terms are 155 and 267; still, there may be some connection to limited height $k$-ary rooted trees.