I think this is more of a confusion of language and nothing else. If $f(x) \in K[x] $ is a specific polynomial then the coefficients of $f$ are nothing but specific members of $K$.
And then if you have a formula for roots of $f$ which involves a combination of some members of $K$ along with operations like $+, -, \times, /, \sqrt[n] {. } $ then the coefficients of $f$ themselves being members of $K$ can not be visually located in the formula. Any member of $K$ can for example easily be written as a combination of any given number of members of $K$ using just the field operations.
You are perhaps trying to think of an example where the coefficients are literals like in case of $x^2+ax+b$ and $K=\mathbb{Q} $, but again this is wrong. In such case the field should be $K=\mathbb{C} (a, b) $.
Let us then assume that we have a literal polynomial $$f(x)= x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$$ over field $K=\mathbb{C} (a_1,a_2,\dots,a_n)$. If $f$ is solvable by radicals over $K$ then formula for roots involves arithmetic operations and radicals (nested if needed) applied on members of $K$ and it does include the literal coefficients of $f$ because they are what $K$ is made of. This is easily seen to be the case in case of quadratic or cubic equations which are known to be solvable.
Thus the coefficients always enter the formula for roots if there is a formula available.
Also note the well known fact (established by Abel well before Galois) that the polynomials with literal coefficients are solvable over their field of coefficients ($K=\mathbb{C} (a_1,a_2,\dots,a_n)$) if and only if $n<5$.
To summarize such an example which you are seeking does not exist.
I have tried to discern the meaning of the comment by reuns and it appears related to the treatment of solvable quintic given by Dummit and Foote in his Abstract Algebra.
They describe a criterion to check whether a given quintic $$f(x) =x^5+ax^4+bx^3+cx^2+dx+e\in\mathbb{Q}[x]$$ is solvable over $\mathbb{C} $. The idea is to form a complicated polynomial of degree 6 in $\mathbb{Q} [x] $ with coefficients made using coefficients of $f$ and checking whether it has a rational root or not.
And if the polynomial of degree 6 mentioned above does have a rational root then $f$ is solvable by radicals over $\mathbb{C} $. You perhaps want to check (for this case) if there is a formula for roots based on elements of $K=\mathbb {C} (a, b, c, d, e) $. I think there is such a formula but I am not sure.
Usually when we consider the problem of solvability of a polynomial $f(x) \in K[x] $, the field $K$ is the smallest field containing the coefficients of $f$. In this case if the polynomial is solvable by radicals over $K$ then the roots can be expressed in terms of coefficients of $f$ via arithmetic operations and radicals.
Enlarging the field $K$ to some extension $L$ and checking solvability over $L$ makes the problem simpler (trivial if $L$ is splitting field of $f$).
Also if we consider the scenario where $f(x) \in K[x] $ is solvable by radicals over $K$ and $F\subset K$ is the smallest field containing the coefficients we need to investigate the problem of solvability of $f$ over $F$ separately and one can't deduce anything from its solvability over $K$.
Thus your problem makes sense only in the usual setting where the solvability is checked over the field of coefficients and then (to repeat what I said earlier) the kind of example you seek does not exist.
Best Answer
Answers to How can I tell if $x^5 - (x^4 + x^3 + x^2 + x^1 + 1)$ is/is not part of the solvable group of polynomials? use a dedicated CAS function (e.g., Maple's $\texttt{galois()}$) to compute that the Galois group $\operatorname{Gal}(f_n)$ of $f_n(x) := x^n - \sum_{i = 0}^{n - 1} x^i$ is the full group $S_n$ for $n \leq 12$. For $n \geq 5$, $S_n$ is not solvable, and so for $5 \leq n \leq 12$ the roots of $f_n$ cannot be expressed in terms of radicals. In particular this applies to the polynomial in this question: $$\color{#df0000}{\boxed{\textrm{The roots of $x^6 - x^5 - x^4 - x^3 - x^2 - x - 1$ cannot be expressed in terms of radicals.}}}$$
To show this result for $f_6$ without using $\mathtt{galois()}$ or a similar function one can proceed as follows.
Proof. First observe that $f_6$ is irreducible modulo $5$, hence irreducible over $\mathbb Z[x]$. Now, recall that
A theorem of Dedekind [pdf] asserts that if $p$ is a prime not dividing the discriminant of a polynomial $f$ and $f(x)$ factors into distinct irreducible polynomials modulo $p$ as $$f(x) \equiv g_1(x) \cdots g_r(x) \pmod p ,$$ then the Galois group $\operatorname{Gal}(f)$ of $f$ (regarded as a subgroup of the symmetric group $S_n$ of permutations of its roots) contains an element of cycle type $(d_1, \ldots, d_r)$, where $d_i := \deg g_i$, $1 \leq i \leq r$.
If a transitive subgroup $G \leq S_n$—in particular, the Galois group of an irreducible polynomial of degree $n$—contains both a transposition and an $(n - 1)$-cycle, then $G = S_n$.
So, it suffices to use Dedekind's Theorem to find primes $p$ not dividing the discriminant $\Delta := 205\,937$ of $f_6$ for which $f_6$ factors modulo $p$ into products of irreducible polynomials of appropriate degree and then apply (2) to conclude that $\operatorname{Gal}(f_6) = S_n$. Since $\Delta$ is prime, the restriction on $p$ is just $p \neq \Delta$.
First, $f_6(x) \equiv (x + 2) g(x) \pmod {17}$ for an irreducible quintic polynomial $g$, so $\operatorname{Gal}(f_6)$ contains a $5$-cycle. On the other hand, $f_6(x) \equiv (x + 2) (x + 4) h(x) \pmod 5$ for an irreducible quartic polynomial $h$, so $\operatorname{Gal}(f_6)$ contains a product a $4$-cycle $\sigma$, and in particular its square $\sigma^2 \in \operatorname{Gal}(f_6)$ is a transposition. So, by (2) $\operatorname{Gal}(f_6) = S_6$. $\blacksquare$
I'm not aware of any way to express roots of a general sextic polynomial beyond simply naming them, but we can apply a clever transformation (courtesy of achille hui in the comments) to write the roots of this particular sextic in terms of a particular infinite series. Computing $(x - 1) f_6(x)$, substituting $x = 2^{-1 / 6} t^{-1}$ and multiplying through by $t^7$ gives the septic polynomial $t^7 - t + 2^{-7 / 6}$. Now, referring to a derivation of Glasser gives that we can write the roots $t_k$, $k = 1, \ldots, 6$, of the polynomial in $t$ (other than $t_0 = 2^{-1 / 6}$, which corresponds to the $x_0 = 1$) as $$t_k = e^{- \pi k i / 3} - \frac{1}{2^{13 / 6} \cdot 3} \sum_{\ell = 0}^\infty \frac{e^{\pi k \ell i / 3}}{2^{7 \ell / 6} (\ell + 1)!} \frac{\Gamma\left(\frac{7 \ell}{6} + 1\right)}{\Gamma\left(\frac{\ell}{6} + 1\right)} .$$ Substituting back gives that the roots of $f_6(x)$ are $x_k = 2^{-1 / 6} t_k^{-1}$. The roots $t_k$ (and hence $x_k$) can also be written in terms of no more than six values of hypergeometric functions, but an explicit expression is too tedious to write out here; for more see the link in this paragraph.
The unique positive root, $1.98358\!\ldots$, of $f_6$ is sometimes called the Hexanacci constant. See also the MathWorld article on Hexanacci numbers, a generalization of the Fibonacci numbers, which can be expressed in terms of the roots of $f_6$.