You want $a+b+c+d = 12$ where $2 \leq a \leq b \leq c \leq d \leq 5$. Let $a_1 = a-2$, $b_1 = b-2$, $c_1 = c-2$ and $d_1 = d-2$. This gives us that $$a_1 + b_1 + c_1 + d_1 = 4$$ where $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
Let $b_1 = a_1 + b_2$, $c_1 = b_1 + c_2$ and $d_1 = c_1 + d_2$. Then we need $$a_1 + (a_1 + b_2) + (a_1 + b_2 + c_2) + (a_1 + b_2 + c_2 + d_2) = 4$$
i.e. $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
where $0 \leq a_1,b_2,c_2,d_2$.
Note that $a = a_1 +2$, $b = a_1 + b_2 + 2$, $c = a_1 + b_2 + c_2 + 2$ and $d = a_1 + b_2 + c_2 + d_2 + 2$
Now all we want is $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
such that $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
This means $a_1 \leq 1$.
If $a_1 = 1$, then $b_2 = c_2 = d_2 = 0$. Hence the solution is $$(a,b,c,d) = (3,3,3,3)$$
If $a_1 = 0$, then $$3b_2 + 2 c_2 + d_2 = 4$$ where $0 \leq b_2,c_2,d_2$.
This means $b_2 \leq 1$.
If $b_2 = 1$, then $c_2 = 0$ and $d_2 = 1$. Hence, the solution is $$(a,b,c,d) = (2,3,3,4)$$
If $b_2 = 0$, then $$2c_2 + d_2 = 4$$ where $0 \leq c_2,d_2$. This gives us $(c_2,d_2) = (2,0)$, $(c_2,d_2) = (1,2)$ and $(c_2,d_2) = (0,4)$. But $d_1 \leq 3$. Hence, the last solution is not possible.
Hence, these now give the solutions
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
Hence, the only four possible solutions for $a+b+c+d = 12$, with the constraint that $a,b,c,d \in \{2,3,4,5\}$ are
$$(a,b,c,d) = (3,3,3,3)$$
$$(a,b,c,d) = (2,3,3,4)$$
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
There are $89$ ways (and $89$ is the $11$th Fibonacci number).
You're looking for "restricted compositions" of $12$ (the restriction being that each part is at least $2$).
In Sage (doc):
sage: Compositions(12, min_part = 2).list()
[[12], [10, 2], [9, 3], [8, 4], [8, 2, 2], [7, 5], [7, 3, 2], [7, 2, 3], [6, 6], [6, 4, 2], [6, 3, 3], [6, 2, 4], [6, 2, 2, 2], [5, 7], [5, 5, 2], [5, 4, 3], [5, 3, 4], [5, 3, 2, 2], [5, 2, 5], [5, 2, 3, 2], [5, 2, 2, 3], [4, 8], [4, 6, 2], [4, 5, 3], [4, 4, 4], [4, 4, 2, 2], [4, 3, 5], [4, 3, 3, 2], [4, 3, 2, 3], [4, 2, 6], [4, 2, 4, 2], [4, 2, 3, 3], [4, 2, 2, 4], [4, 2, 2, 2, 2], [3, 9], [3, 7, 2], [3, 6, 3], [3, 5, 4], [3, 5, 2, 2], [3, 4, 5], [3, 4, 3, 2], [3, 4, 2, 3], [3, 3, 6], [3, 3, 4, 2], [3, 3, 3, 3], [3, 3, 2, 4], [3, 3, 2, 2, 2], [3, 2, 7], [3, 2, 5, 2], [3, 2, 4, 3], [3, 2, 3, 4], [3, 2, 3, 2, 2], [3, 2, 2, 5], [3, 2, 2, 3, 2], [3, 2, 2, 2, 3], [2, 10], [2, 8, 2], [2, 7, 3], [2, 6, 4], [2, 6, 2, 2], [2, 5, 5], [2, 5, 3, 2], [2, 5, 2, 3], [2, 4, 6], [2, 4, 4, 2], [2, 4, 3, 3], [2, 4, 2, 4], [2, 4, 2, 2, 2], [2, 3, 7], [2, 3, 5, 2], [2, 3, 4, 3], [2, 3, 3, 4], [2, 3, 3, 2, 2], [2, 3, 2, 5], [2, 3, 2, 3, 2], [2, 3, 2, 2, 3], [2, 2, 8], [2, 2, 6, 2], [2, 2, 5, 3], [2, 2, 4, 4], [2, 2, 4, 2, 2], [2, 2, 3, 5], [2, 2, 3, 3, 2], [2, 2, 3, 2, 3], [2, 2, 2, 6], [2, 2, 2, 4, 2], [2, 2, 2, 3, 3], [2, 2, 2, 2, 4], [2, 2, 2, 2, 2, 2]]
sage: Compositions(12, min_part = 2).cardinality()
89
The number for general $n$ in place of $12$ would be interesting to calculate. Using Sage again:
sage: print ','.join(str(Compositions(n, min_part = 2).cardinality()) for n in range(20))
1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584
Searching for that on OEIS gives A212804 as the sequence.
This of course is the Fibonacci sequence with an additional "1" at the beginning.
Now it would be interesting to prove that it is so for general $n$. This is easy now that we know what to prove. Let's call compositions with each part at least $2$ as "good" compositions. Let $C_n$ be the number of "good" compositions of $n$. Any such "good" composition either begins with $2$ or with a number greater than 2:
If it starts with $2$, then it can be got by prepending a $2$ to a "good" composition of $n-2$ (and there are $C_{n-2}$ of them)
If it starts with a number greater than $2$, then it can be got by taking a "good" composition of $n-1$ (there are $C_{n-1}$ of them) and incrementing the first element by one.
So $C_n = C_{n-2} + C_{n-1}$, which along with the initial conditions $C_0 = 1$ and $C_1 = 0$, proves what we wanted to prove.
Best Answer
These are called integer partitions, and there are a lot of fun combinatorics that goes along with them. If we let $p(n)$ denote the number of integer partitions of $n$, the generating function is given by
$$ \sum_{n=0}^{\infty} p(n) x^n = \prod_{j=1}^{\infty}\frac{1}{1-x^j} $$
it's not too hard to convince yourself this is correct: the product on the right can be seen as
$$(1+x^{1\cdot1}+x^{2\cdot1}+x^{3 \cdot 1}+\dots)(1+x^{1\cdot2}+x^{2\cdot2}+x^{3\cdot 2}+\dots)(1+x^{1\cdot3}+x^{2\cdot3}+x^{3\cdot3}+\dots)\dots$$
so for example if we look at how one might get $x^5$:
and so $p(5) = 7$, as you enumerated.
Unfortunately there's no nice formula for this, but asymptotically we have
$$p(n) \sim \frac{e^{\sqrt n \,c}}{4\sqrt 3 \,n},$$
where $ c = \sqrt \frac{2}{3} \pi$. This is harder to prove, but I believe it is due to Hardy and Ramanujan.
Here is the most straightforward proof I could find, though it uses some results from Complex Analysis.
This is a proof using only real analysis, but the result is slightly weaker.