Not an answer, but the formulae are too bulky for a comment.

After a lot of experimenting I found a formula for the coefficients of $\phi_{n}(x^{-1})$. (Currently without formal proof, but the evidence is really strong.)

I conjecture that $\phi_{n}(x^{-1})$ has the following form
$$
\phi_{n}(x^{-1})= (-1)^{n} + \sum_{k=2}^{2 n} c_{n,k} x^{k},
$$
where
$$
c_{n,k}=c^{<}_{n,k}:=\frac{(-1)^{n+1} n!}{(n-k+1)!}\  {}_{2}F_{3}(1-\frac{k}{2},\frac{3}{2}-\frac{k}{2};2,2-k,n-k+2;4)
$$
for $2\leq k \leq n $ and
$$
c_{n,k}=c^{>}_{n,k}:=\frac{(-1)^{k} n!}{(k-n)!}{{n-1}\choose{k-n-1}}  {}_{2}F_{3}(\frac{k}{2}-n,\frac{k}{2}-n+\frac{1}{2};1-n,k-n,k-n+1;4)
$$
for $n< k \leq 2 n $. The ${}_{2}F_{3}$ are generalized hypergeometric functions.

The series of the hypergeometric functions in the formula for $c^{<}_{n,k}$ terminates, since one of the nominatorial parameters of the hypergeometric  is always a non-positive integer.

The sum of the hypergeometric functions in the formula for $c^{>}_{n,k}$ terminates also. The non-positive denominatorial parameter, $n-1$,  is always overruled by one of the nominatorial parameters, since $1-n<\frac{k}{2}-n$ for the relevant values. In cases of equality of the nominatorial and denominatorial parameters they cancel and the function collapses to a well defined ${}_{1}F_{2}$. This overruling and cancellation has to be carefully implemented in math algebra systems like Mathematica, since it is not easily recognized by them.

The sign of $c^{>}_{n,k}$ alternates depending on $n$ and $k$ as already mentioned in other answers.

The formulae were found by using $f$ with a decoration of the variable $x$
$$
f(x)=\exp( - x \ z - \frac{y}{x} )
$$
and expand the result for $\phi$ in $x$, $z$, and $y$ using Mathematica. The formulae for the coefficients of the resulting multinomials were easily guessed and after setting $z=1$ and $y=1$ only Mathematica solvable sums remained.