let's assume $p[n]$ is the name of this partitioning method Let's see some examples:
$n=3$:
all possibilities are: $[(3,0),(2,1),(1,1,1)]$
all cases don't meet the condition $minSize > 1$
so $p[3]=0$.
$n=4$:
all possibilities are: $[(4,0), (3,1), (2,2), (2,1,1), (1,1,1,1)]$
only the case $(2,2)$ stands the condition $minSize>1$, now, let's see how many ways can we split 4 into 2 groups of 2 elements $(2,2)=3$
so $p[4]=(2,2)=3$
$n=5$:
all possibilities are $[(5,0),(4,1),(3,1,1),(3,2),(2,2,1),(2,1,1,1),(1,1,1,1,1)]$
only the case $(3,2)$ stands the condition $minSize>1$, $(3,2)=10$ (it starts to get harder to compute it with pen!)
so $p[5]=(3,2)=10$
$p[6]=(4,2)+(3,3)+(2,2,2)=???$
$p[7]=(4,3)+(3,2,2)=???$
$p[8]=(6,2)+(5,3)+(4,4)+(4,2,2)+(3,3,2)+(2,2,2,2)=???$
and it get harder and harder, can you see the logic?
EDIT
My most serious trials so far is to try to customize Stirling numbers of the second kind to solve this problem, but still have a lot of issues in formalizing it, and I am not sure if it is correct to use it, it is just my guess.
P.S: Feel free to add tags to this question, I am not sure how to tag it correctly