To find a closed form for $f_X(m)$ in terms of the $S_\ell$ we first
require the exponential formula for the cycle index of the unlabeled set
operator
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}.$$
Let $A$ be a generating function in some number of
variables and let $B = c_B X_B$ a contributing monomial term where
$c_B$ is the leading coefficient (positive) and $X_B$ the product
of the variables to their respective powers, so that $A= \sum_{B\in A}
c_B X_B$. We then have from first principles that the generating
function of sets drawn from $A$ containing $m$ elements is
$$[z^m] \prod_{B\in A} (1 + z X_B)^{c_B}.$$
Manipulate this to obtain
$$[z^m] \prod_{B\in A} \exp \log (1 + z X_B)^{c_B}
\\ = [z^m] \prod_{B\in A}
\exp \left(- c_B \log \frac{1}{1 + z X_B} \right)
\\ = [z^m] \exp
\sum_{B\in A} \left(- c_B \log \frac{1}{1 + z X_B} \right)
\\ = [z^m] \exp
\sum_{B\in A} \left(- c_B \sum_{\ell\ge 1} (-1)^\ell X_B^\ell
\frac{z^\ell}{\ell} \right)
\\ = [z^m] \exp \left( \sum_{\ell\ge 1} (-1)^{\ell-1}
\left(\sum_{B\in A} c_B X_B^\ell\right)
\frac{z^\ell}{\ell} \right).$$
But $\sum_{B\in A} c_B X_B^\ell$ is by definition the Polya substitution
applied to $A$ through the cycle index variable $a_\ell$ and we have
proved the exponential formula for the unlabeled set operator, which says
that
$$Z(P_m) = [z^m] \exp
\left(\sum_{\ell\ge 1} (-1)^{\ell-1}
a_\ell \frac{z^\ell}{\ell}\right).$$
Now we have from the definition applying PET that
$$f_X(m) = Z(P_m; x_1+x_2+\cdots+x_n).$$
Hence
$$f_X(m) = [z^m] \exp
\left(\sum_{\ell\ge 1} (-1)^{\ell-1}
S_\ell \frac{z^\ell}{\ell}\right)
\\ = [z^m] \prod_{\ell\ge 1}
\exp \left( (-1)^{\ell-1}
S_\ell \frac{z^\ell}{\ell}\right)
\\ = [z^m] \prod_{\ell\ge 1}
\sum_{q\ge 0} \frac{1}{q!} (-1)^{q(\ell-1)}
S_\ell^q \frac{z^{\ell q}}{\ell^q}.$$
We want to expand this product. We are interested in the coefficient
on $[z^m]$ so we consider integer partitions $\lambda\vdash m$. Let
the partition be $1^{q_1} 2^{q_2} 3^{q_3} \cdots$ where all but a
finite number of exponents are zero. We now obtain
$$[z^m] \sum_{\lambda\vdash m} \prod_{\ell\ge 1}
\frac{1}{q_\ell!}
(-1)^{(\ell-1)q_\ell} S_\ell^{q_\ell}
\frac{z^{\ell q_\ell}}{\ell^{q_\ell}}.$$
Since $\sum_{\ell\ge 1} \ell q_\ell = m$ this becomes
$$\sum_{\lambda\vdash m} \prod_{\ell\ge 1}
\frac{1}{q_\ell!}
(-1)^{(\ell-1)q_\ell} S_\ell^{q_\ell}
\frac{1}{\ell^{q_\ell}}
\\ = (-1)^m \sum_{\lambda\vdash m}
(-1)^{\sum_{\ell\ge 1} q_\ell}
\prod_{\ell\ge 1} S_\ell^{q_\ell}
\frac{1}{q_\ell! \times \ell^{q_\ell}}.$$
We thus get the closed form
$$\bbox[5px,border:2px solid #00A000]{
f_X(m) = \frac{(-1)^m}{m!} \sum_{\lambda\vdash m}
(-1)^{\sum_{\ell\ge 1} q_\ell}
\left(\prod_{\ell\ge 1} S_\ell^{q_\ell} \right)
\left(m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}\right).}$$
This yields for example
$$f_X(4) =
\frac{1}{4!}
\left({S_{{1}}}^{4}-6\,{S_{{1}}}^{2}S_{{2}}
+8\,S_{{1}}S_{{3}}+3\,{S_{{2}}}^{2}-6\,S_{{4}}\right)$$
and
$$f_X(5) =
\frac{1}{5!}
\left({S_{{1}}}^{5}-10\,{S_{{1}}}^{3}S_{{2}}+20\,{S_{{1}}}^{2}S_{{3}}
+15\,S_{{1}}{S_{{2}}}^{2}-30\,S_{{1}}S_{{4}}
-20\,S_{{2}}S_{{3}}+24\,S_{{5}}\right).$$
We now show that
$$m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}$$
counts the number of permutations with cycle structure $\lambda.$
First, selecting the values to go on the cycles yields the multinomial
coefficient
$$\frac{m!}{\prod_{\ell\ge 1} (\ell!)^{q_\ell}}.$$
A set of $\ell$ values gives $\frac{\ell!}{\ell}$ cycles:
$$\prod_{\ell\ge 1} \left( \frac{\ell!}{\ell} \right)^{q_\ell}.$$
Any permutation of the cycles of length $\ell$ yields the same
permutation:
$$\prod_{\ell\ge 1} \frac{1}{q_\ell!}.$$
Multiply these to obtain
$$\frac{m!}{\prod_{\ell\ge 1} (\ell!)^{q_\ell}}
\prod_{\ell\ge 1} \left( \frac{\ell!}{\ell} \right)^{q_\ell}
\prod_{\ell\ge 1} \frac{1}{q_\ell!}
= m! \prod_{\ell\ge 1} \frac{1}{q_\ell! \times\ell^{q_\ell}}.$$
This is the claim and concludes the argument.
As an addendum we have by inspection for the boxed closed form that
it is given by a substitution into $Z(Q_m)$, the cycle index of the
symmetric group (the variable $S$ is in use already), which is
$$Z(Q_m; a_\ell = (-1)^{\ell -1} S_\ell).$$
With $Z(Q_m)$ being the average of all $m!$ permutations factorized
into cycles, where $a_\ell$ stands for a cycle of $\ell$ elements we have
$$Z(Q_m) =
\frac{1}{m!} \sum_{\lambda\vdash m}
\left(\prod_{\ell\ge 1} a_\ell^{q_\ell} \right)
\left(m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}\right).$$
Now put $a_\ell = (-1)^{\ell-1} S_\ell$ to get the boxed form. The
substitution $a_\ell := (-1)^{\ell-1} a_\ell$ converts the unlabeled
multiset operator into the unlabeled set operator through their cycle
indices.
There is also some Maple code for the curious who would want to study
these polynomials and verify the correctness of the closed form.
with(combinat);
fX1 :=
proc(m, n)
local t;
coeff(expand(mul(1+u*x[t], t=1..n)), u, m);
end;
fX2 :=
proc(m)
local res, part, mset, ent;
res := 0;
part := firstpart(m);
while type(part, list) do
mset := convert(part, `multiset`);
res := res +
(-1)^add(ent[2], ent in mset)
* mul(S[ent[1]]^ent[2], ent in mset)
* m!/mul(ent[2]!*ent[1]^ent[2], ent in mset);
part := nextpart(part);
od;
res*(-1)^m/m!;
end;
fX :=
proc(m, n)
local l, sl, t;
sl := [seq(S[l] = add(x[t]^l, t=1..n), l=1..m)];
expand(subs(sl, fX2(m)));
end;
pet_cycleind_symm :=
proc(n)
local l;
option remember;
if n=0 then return 1; fi;
expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
fX2A :=
proc(m)
local sl, l;
sl := [seq(a[l] = (-1)^(l-1) * S[l], l=1..m)];
subs(sl, pet_cycleind_symm(m));
end;
