There is a nice computation here that uses the Polya Enumeration
Theorem to treat the case when order is not important, i.e. when
each multiset of choices only contributes once. Applying PET we get
the species equation
$$\mathfrak{M}\left(\sum_{q=0}^{n-1} \mathcal{U}^{2^q}\right).$$
This immediately gives the ordinary generating function
$$F(u) = Z(S_n)\left(\sum_{q=0}^{n-1} u^{2^q}\right)$$
where $Z(S_n)$ is the cycle index of the symmetric group and the
second pair of parenthesis denote cycle index substitution.
Digress for a moment to consider the labelled species $\mathcal{Q}$
given by the specification
$$\mathcal{Q} =
\mathfrak{P}(\mathcal{A}_1 \mathfrak{C}_{=1}(\mathcal{Z})
+ \mathcal{A}_2 \mathfrak{C}_{=2}(\mathcal{Z})
+ \mathcal{A}_3 \mathfrak{C}_{=3}(\mathcal{Z})
+ \cdots).$$
This corresponds to the generating function
$$Q(z) =
\exp\left(a_1 z +
a_2 \frac{z^2}{2} + a_3 \frac{z^3}{3} + \cdots\right).$$
We then have by inspection the result that $Q(z)$ is the OGF of the
cycle index of the symmetric group:
$$Z(S_n) = [z^n] Q(z).$$
Returning to $F(u)$ we get
$$F(u) = [z^n] \exp
\left(\sum_{k\ge 1} \frac{z^k}{k} \sum_{q=0}^{n-1} u^{k2^q} \right).$$
We will not pick up any large cycles of length $m$ greater than $n$
from the species since the term $z^m/m$ will prevent these from
appearing in $[z^n] Q(z).$
Now to conclude note that we are only interested in $F(1)$ which gives
the total number of different sums that appear and disregards their
value. Observe that
$$\left.\sum_{q=0}^{n-1} u^{k2^q}\right|_{u=1} = n$$ so that
$$F(1) = [z^n] \exp\left(n \sum_{k\ge 1} \frac{z^k}{k}\right)
= [z^n] \exp\left(n \log\frac{1}{1-z}\right)
= [z^n] \left(\frac{1}{1-z}\right)^n \\=
{n+n-1\choose n-1} = {2n-1\choose n}.$$
We see that the removal of the dependence on the actual value of the
sum has converted the problem into an application of stars-and-bars
where the value $p$ at position $k$ indicates that $p$ copies of
element at position $k$ in the source vector were chosen.
Of course we have the generating functions. E.g. for $n=3$ we get
$${u}^{12}+{u}^{10}+{u}^{9}+{u}^{8}+{u}^{7}+2\,{u}^{6}+{u}^{5}+{
z}^{4}+{u}^{3}$$
and for $n=5$ we have
$${u}^{80}+{u}^{72}+{u}^{68}+{u}^{66}+{u}^{65}+{u}^{64}+{u}^{60}
+{u}^{58}+{u}^{57}+2\,{u}^{56}+{u}^{54}+{u}^{53}\\+2\,{u}^{52}+{
z}^{51}+2\,{u}^{50}+{u}^{49}+2\,{u}^{48}+{u}^{46}+{u}^{45}+3\,
{u}^{44}+{u}^{43}+3\,{u}^{42}+2\,{u}^{41}\\+3\,{u}^{40}+{u}^{39}
+3\,{u}^{38}+2\,{u}^{37}+4\,{u}^{36}+2\,{u}^{35}+3\,{u}^{34}+2
\,{u}^{33}+3\,{u}^{32}+{u}^{31}+4\,{u}^{30}\\+3\,{u}^{29}+4\,{u}
^{28}+3\,{u}^{27}+4\,{u}^{26}+2\,{u}^{25}+4\,{u}^{24}+3\,{u}^{
23}+4\,{u}^{22}+3\,{u}^{21}+4\,{u}^{20}\\+2\,{u}^{19}+3\,{u}^{18
}+2\,{u}^{17}+3\,{u}^{16}+3\,{u}^{15}+3\,{u}^{14}+2\,{u}^{13}+
3\,{u}^{12}+2\,{u}^{11}+2\,{u}^{10}\\+2\,{u}^{9}+2\,{u}^{8}+{u}^
{7}+{u}^{6}+{u}^{5}.$$
The reader may want to verify some of these. E.g. the three multisets for the sum $15$ are $1+1+1+4+8, 1+2+2+2+8$ and $1+2+4+4+4.$
