Note that to find the total number of possible subsets of $n$ distinct elements, it is the same as putting the $n$ distinct elements into $2$ distinct boxes. In other words, each distinct element is either in the subset or not in the subset.
Therefore, the total number of subsets = $2^n$ (this includes the $\varnothing$ of no elements)
At the same time, the total number of subsets can also be calculated as you suggested in $(1)$ and $(2)$: $$\sum_{k=0}^{n}\frac{n!}{(n-k)!k!} = \sum_{k=0}^{n}\binom{n}{k}=2^n$$
This will allow you to find a "nicer form" for your expression in $(2)$.
$$ \binom{n}{0}+\binom{n}{1}+\binom{n}{n}+\sum_{k=2}^{n-1}\binom{n}{k}=2^n$$
$$\sum_{k=2}^{n-1}\binom{n}{k}=2^{n}-n-2$$
When there are identical terms, we will have to first count the number of subsets of the distinct terms (let us assume there are $a$ number of some identical term).
We will have $2^{(n-a)}$ subsets of distinct terms.
Now we will count the number of subsets of $a$ identical terms or find the number of ways to distribute $a$ identical terms into $2$ distinct boxes:
$$\binom{a+2-1}{2-1} = a+1$$
This uses the "stars and bars" method where the distribution of $n$ identical objects into $k$ distinct boxes, empty boxes are allowed is given by:
$$\binom{n+k-1}{k-1}$$
In the above case, we had $a$ objects and $2$ distinct boxes, i.e, "in the subset" or "not in the subset".
We can then multiply the two together to find the number of subsets of the $(n-a)$ distinct terms and $a$ identical terms again including $\varnothing$. Once you get your final value, it only includes the null set once so you can choose to minus $1$ if you do not wish to count the null set.
If we have multiple sets of identical elements, for example, $S=\{el_1,el_1,el_2,el_3,...,el_n\}$ and with $a$ number of '$el_a$'s, $b$ number of '$el_b$'s and so on.
The total number of subsets is then given by:
$$2^{(n-a-b-\dots)} (a+1)(b+1)(c+1)\dots$$
Let $o$ and $e$ be the number of odd and even numbers in $S$, respectively, with $o+e=N$. If $o=0$, all $2^N$ subsets have even sum. If $o\ne0$, then half of the $2^o$ subsets of the odd numbers contain an even number of odd numbers and hence have an even sum, and they can be combined with all $2^e$ subsets of the even numbers, for a total of $2^{e+o-1}=2^{N-1}$ subsets.
Best Answer
The example suggests you are concerned with forming sums, possibly with repeated summands, by drawing from a multiset of positive integers. In this respect you are asking about the size of the feasible set of an integer programming problem, which is (without other assumptions) NP-hard.
This is an interesting problem, one that has been discussed previously at MathOverflow and at StackOverflow (but not on Math.SE as far as I could tell).
Let me offer an argument why we should not expect a straightforward useful formula for these counts, at least not one that can be evaluated in polynomial time. Let us restrict attention for the moment to sets of $N$ distinct values (no repetitions, so a strict set and not a multiset is involved).
If there were a "simple" (polynomial time in size of inputs) way to determine how many subset-sums occur, then we would certainly know, by comparing with the maximum possible value $2^N$, whether two distinct subsets give the same sum. This is a decision problem introduced by Woeginger and Yu (1992) in "On the equal-subset-sum problem", which they showed to be NP-complete. Therefore no formula evaluable in polynomial time is possible, unless P=NP. [A more recent paper by Cieliebak et al (2008) summarizes the complexity of several variants of equal-subset-sum, and moreover it is not behind a "paywall".]
Below I sketch an algorithm more sophisticated than simple "brute force" consideration of all possible $2^N$ (multi)subsets. I will then add an example and compare my suggestion to the interesting dynamic programming approach sketched in an answer to the StackOverflow post linked above.
One way to count the possible sums is to construct them all systematically by how many summands are used (counting the allowed repetitions afforded by the multiset). That is, the empty subset gives a zero sum. The single summand sums are precisely the distinct elements of the multiset. The distinct sums do not vary with the order of the summands, so we can systematically construct the distinct sums of $n$ summmands by extracting the distinct $n$-multisubsets of the given multiset in (say) lexicographic order.
An algorithm for doing so was described in this recent Answer.
It is then necessary to redact those $n$-summand sums which agree with results obtained by the same or fewer number of summands, and merge those which are novel. We will eliminate duplicates most efficiently if we maintain the list of sums achieved in order. While (as you note) lexicographic generation does not always produce new sums in ascending order, it does tend to produce long runs of ascending order.
The "dynamic programming approach" sketched in the SO Answer alluded to above proceeds instead by recursively including the available elements from smallest to largest. We only need bits up to the sum of all elements to record which of these respective sums are achieved or not. This is a pseudo-polynomial time (and space) algorithm.
Example (to do)