This is not a complete answer, but summarizes three provable cases. For simplicity (and better familiarity with existing posts), let $f_n(x)=x^{n-1}+2x^{n-2}+\dots+(n-1)x+n$.
Case 1: $n+1$ is a prime. As you have noticed, in many cases we can use Eisenstein criterion for $f_n(x+1)$. We can show that this will work when $n+1$ is a prime. We have
$$
f_n(x+1)=\sum_{i=0}^{n-1}(n-i)(x+1)^i=\sum_{i=0}^{n-1}(n-i)\sum_{j=0}^{i}\binom{i}{j}x^j=\sum_{j=0}^{n-1}\sum_{i=j}^{n-1}(n-i)\binom{i}{j}x^j
$$
and the coefficient at $x^j$ is
$$[x^j]f_n(x+1)=\sum_{i=j}^{n-1}(n-i)\binom{i}{j}=\sum_{i=0}^{n-1}(n-i)\binom{i}{j}=\binom{n+1}{j+2}$$
Now having $p=n+1$ a prime, we have $p \mid \binom{n+1}{j+2}$ for $j+2<n+1$, i.e. $j<n-1$. Also for $j=0$ the coefficient becomes $\binom{n+1}{2}=\frac{n}{2}(n+1)$ and it is clearly divisible by $p$ but not by $p^2$, so the polynomial satisfies conditions of Eisenstein criterion for $p$ and is therefore irreducible.
Case 2: $n$ is a prime.
The irreducibility can be also shown for $n$ being a prime as discussed in comments. The idea is that all complex roots of the $f_n(x)$ lie outside of the unit circle in a complex plane, which can be shown by looking at roots of $(x-1)f_n(x)=x^n+x^{n-1}+\dots+x-n$. Then having $f_n(x)=p(x)q(x)$ with $n$ prime implies that without loss of generality we can take $p(x)$ such that $|p(0)|=1$. But $|p(0)|=\prod |x_k|>1$ gives a contradiction. More details on this can be found here Explain proof of irreducibility of $x^{p-1} + 2x^{p-2}+ \dots +(p-1)x + p$.
Case 3: $n$ is a prime power.
The previous case argument can be modified to work on any prime power. Say we have $n=p^k$, and $f_n(x)=g(x)h(x)$ where coefficients of $g(x),h(x)$ are $a_i$,$b_i$ respectively. For constant coeficient we have $n=p^k=a_0b_0$, so $a_0$ and $b_0$ are non-negative powers of $p$. Then for linear coefficient of $f_n(x)$ we have $n-1=a_1b_0+a_0b_1$. Since $(n-1,n)=1$, one of $a_0,b_0$ must be equal $\pm 1$, say $a_0$. Finally, again as in previous case we reach contradiction by looking at $|g(0)|=\prod |x_k|>1$.
Remaining cases are problematic and I could not find anything about this in literature or prove it. I've tried to get more by looking at the roots but it did not get me much, here is at least an observation I stumbled upon - all complex roots of the polynomial lie in annulus $1<|x_k|<\sqrt[n]{2n+1}$. First inequality is already proven above. For the second inequality notice that $(x-1)^2f_n(x)=x^{n+1}-(n+1)x+n$, and so for a root $z$ we have
$$n=|z^{n+1}-(n+1)z|=|z||z^n-(n+1)|.$$
Now being interested only in roots $|z|>1$, this gives $n>|z^n-(n+1)|$, which geometrically means that $z^n$ distance from $n+1$ is less than $n$, so especially $|z^n|<n+1+n$, from which we have $|z|<\sqrt[n]{2n+1}$.
Hopefully someone in the future will have an idea on the remaining cases, but notice that this problem was posted in Putnam and there it was only for $n$ being prime, so this seems like a hard problem.
Update: Literature reference
Okay I have managed to find a reference related to the problem, specifically see Classes of polynomials having only one
non-cyclotomic irreducible factor by A. Borisov, M. Filaseta,
T. Y. Lam and O. Trifonov (http://matwbn.icm.edu.pl/ksiazki/aa/aa90/aa9023.pdf). It states a conjecture about $f(x)=1+x+x^2+\dots+x^n$, then $f'(x)=1+2x+\dots+nx^{n-1}$ being irreducible for all $n \geq 2$, but that is just a reciprocal of $f_n(x)$, so it is equivalent to this problem. It states that the irreducibility has been proven previously for $n=p-1$ or $n=p^r$, and furthermore also when $n+1$ is squarefree or $n=2p-1$. Then the paper itself claims additional results, specifically:
Theorem 1. Let $\varepsilon > 0$. For all but $O(t^{1/3+\varepsilon})$ positive integers $n \leq t$, the derivative of the polynomial $f(x)=1+x+x^2+\dots+x^n$ is irreducible.
Best Answer
This answer does not fully address the question of how to compute an explicit formula from the recursion relation (though it indicates a reference which contains a starting point), and so is not a proper answer, but it is too long for a comment, and I expect it might still be useful.
It follows from a comment by Vladimir Kruchinin in OEIS A042977, "Triangle $T(n,k)$ read by rows: coefficients of a polynomial sequence occurring when calculating the $n$th derivative of Lambert function $W$" that $$\color{#df0000}{\boxed{P_n(w) = \sum_{m = 0}^n \left[ \sum_{j=0}^m {2 n + 1 \choose m - j} \sum_{k=0}^j (-1)^k \frac{(n+k+1)^{n+j}}{(j-k)! k!} \right] w^m}} .$$ See under the heading $\texttt{FORMULA}$, and NB that the expression in this answer differs from the one in OEIS entry by a factor of $(-1)^n$ to account for the formulation in the question. This expression can be deduced from Example 4.3 in Kruchinin's preprint "Derivation of Bell Polynomials of the Second Kind".
In particular, to compute the constant term of $P_n(w)$, we can evaluate the inner sum for $j = 0$. Both summations have only a single term, and it simplifies to the conjectured formula, $$\color{#df0000}{\boxed{P_n(0) = (n + 1)^n}} .$$