You're looking for triples of three numbers (which I'm going to list in nondecreasing order) that sum to 36, and where each one is at least 5.
Instead you can look for triples of three numbers $p \le q \le r$ that sum to $36 - 15 = 21$, where each is nonnegative, and then the numbers you originally sought were $p+5, q+5, r+5$. (i.e., there's a 1-1 correspondence between the collection of partition-sizes that you seek and the collection of $(p, q, r)$ with $0 \le p \le q \le r$ and $p+q+r = 21$.)
These latter triples can be divided into those with $p = 0$, where the remaining numbers sum to 21; those where $p = 1$, and the remaining numbers sum to $20$...but both are at least $1$, those where $p = 2$, and the remaining numbers sum to $19$, but both are at least $2$, etc.
But the same trick applies to those: pairs of numbers, in nondecreasing order, summing to to $19$, but both being at least $2$ have the same count as pairs of (nondecreasing) numbers, summing to $15$, but both being at least $0$.
Let's write
$$
C(n, k)
$$
for the number of $n$-element sets of nonnegative numbers, in increasing order, that sum to exactly $k$. Then what I've said above shows that the number you're looking for is $C(3, 21)$, and that
$$
C(3, 21) = C(2, 21) + C(2, 21-3\cdot 1) + C(2, 21-3\cdot 2) + \ldots + C(2, 21-3\cdot 7)
$$
Now how large is
$C(2, k)$? It, too, satisfies a recurrence: the first number is either $0$ (in which case the remaining number must sum to 21), or it's 1, in which case the remaining number must be at least 1 and sum to 20, i.e., the count of which is the same as numbers that are at least 0 and sum to 19, etc.
$$
C(2, k) = C(1, k) + C(1, k-2 \cdot 1) + C(1, k-2 \cdot 2) + \ldots + C(1, k-2 \cdot (k/2))
$$
where the $k/2$ in the last term should be rounded down [i.e., if $k$ is odd, then we end with $C(1, 1)$ rather than $C(1, -1)$. ]
Now how large is $C(1, s)$ for any nonnegative $s$? It's exactly $1$.
That means that $C(2, k)$ is exactly $\lfloor \frac{k+1}{2} \rfloor$.
So
$$
C(3, 21) = \lfloor \frac{22}{2} \rfloor + \lfloor \frac{19}{2} \rfloor + \lfloor \frac{16}{2} \rfloor + \lfloor \frac{13}{2} \rfloor + \lfloor \frac{10}{2} \rfloor + \lfloor \frac{7}{2} \rfloor + \lfloor \frac{4}{2} \rfloor + \lfloor \frac{1}{2} \rfloor \\
= 11+9 + 8 + 7 + 5 + 3 + 2 = 45.
$$
This seems surprisingly low to me, but I don't see any obvious error (except that my "round down $(k+1)/2$" answer for $C(1, k)$ could be off by one), so I'm going to go ahead and submit it as an answer, and if not an answer, at least a suggested path for you to follow in getting to the correct answer.
Let me just sanity check by writing them down...there aren't that many.
12, 12, 12
11, 12, 13
11, 11, 14
10, 13, 13
10, 12, 14
10, 11, 15
10, 10, 16
9, 13, 14
9, 12, 15
9, 11, 16
9, 10, 17
9, 9, 18
8, 14, 14
8, 13, 15
8, 12, 16
8, 11, 17
8, 10, 18
8, 9, 19
8, 8, 20
7, 14, 15
7, 13, 16
7, 12, 17
7, 11, 18
7, 10, 19
7, 9, 20
7, 8, 21
7, 7, 22
6, 15, 15
6, 14, 16
6, 13, 17
6, 12, 18
6, 11, 19
6, 10, 20
6, 9, 21
6, 8, 22
6, 7, 23
6, 6, 24
Hunh. I seem to get 38. So I probably DID have an off-by-one error in my formula for $C(2, k)$. Anyhow, the answer is 38, and you can work out my off-by-one error by back-tracing, if it pleases you do to so.
Best Answer
The bivariate generating function of the Stirling numbers of the second kind is given by $$G(z,u) = \exp(u(\exp(z)-1))$$ so that $$n! [z^n][u^k] G(z, u) = {n \brace k}.$$ If you want to forbid subsets of $r$ elements, mark them and the generating function becomes $$H(z,u) = \exp\left(uv\frac{z^r}{r!} - u \frac{z^r}{r!} + u(\exp(z)-1)\right) = \exp\left(uv\frac{z^r}{r!} - u \frac{z^r}{r!}\right)G(z,u).$$ This has the property that the coefficient of $[v^0]$ is the generating function of the partitions with sets of size $r$ not allowed, which is $$[v^0] H(z, u) = [v^0] \exp\left(uv\frac{z^r}{r!} \right) \exp\left(-u\frac{z^r}{r!} \right) G(z,u) \\= \exp\left(-u\frac{z^r}{r!} \right) G(z,u) .$$ Now extracting the coefficient of $[z^n]$ from this we find $$n! \sum_{q=0}^{\lfloor n/r\rfloor} \frac{1}{q!} \left(-\frac{u}{r!}\right)^q [z^{n-qr}] G(z, u) .$$ We then obtain for the coefficient of $[u^k]$ the formula $$n! \sum_{q=0}^{\min(k, \lfloor n/r\rfloor)} \frac{(-1)^q}{q!\times (r!)^q} [u^{k-q}] [z^{n-qr}] G(z, u) \\= n! \sum_{q=0}^{\min(k, \lfloor n/r\rfloor)} \frac{1}{(n-qr)!} \frac{(-1)^q}{q!\times (r!)^q} [u^{k-q}] (n-qr)! [z^{n-qr}] G(z, u) \\ = n! \sum_{q=0}^{\min(k, \lfloor n/r\rfloor)} \frac{1}{(n-qr)!} \frac{(-1)^q}{q!\times(r!)^q} {n-qr \brace k-q}.$$ This formula gives the number of set partitions of $n$ elements into $k$ sets with subsets of size $r$ not allowed. Subtract this from ${n\brace k}$ to get the number of partitions where there is at least one subset of $r$ elements.
The OEIS has the case of $k=2$ and $r=1$, the case of $k=3$ and $r=1$ and the case of $k=4$ and $r=1$.
Observation. There is an interesting sanity check here, namely what happens when $r>n$. In this case the formula should produce ordinary Stirling numbers because forbidding subsets of a size larger than the total count of available elements does not affect the statistic. And indeed we have $\lfloor n/r \rfloor = 0$ so that only $q=0$ contributes, giving $$n! \frac{1}{n!} \frac{1}{1\times(r!)^0} {n\brace k} = {n\brace k},$$ so the check goes through and we get the correct value.