I'll give you a yes and a no, depending on exactly what you are looking for:
Yes.
See Sturmfels' paper "Solving Algebraic Equations in Terms of $A$-Hypergeometric Series"
Consider a root of $x_0 + x_1 t + \cdots + x_n t^n$ as a function $F(x_0, x_1, \ldots, x_n)$. Then $F$ is an A-hypergeometric function (also known as a GKZ-hypergeometric function), associated to the $A$-matrix
$$\begin{pmatrix}
0 & 1 & 2 & \cdots & n \\ 1 & 1 & 1 & \cdots & 1
\end{pmatrix}.$$
This result was in some sense known before, but Sturmfels writes these functions down very explicitly.
See Cattani's lectures for a quick intro to A-hypergeometric functions, including the definitions of the terms used above. This fact is Remark 3.23, and Cattani gives the relevant history.
I must admit that I never learned how to translate the modern A-hypergeometric language into the classical Gauss vocabulary. Note, however, that A-hypergeometric functions are power series in many variables, not one. And that brings me to:
No.
Abhyankar has a paper which I wish I understood better; I tried to explain my partial understanding here. Let $F(x_5, x_6)$ be a root of $t^6+x_5 t + x_6=0$. I believe that Abhyankar is showing that $F$ can not be expressed in terms of field operations and holomorphic functions of single variables; it intrinsically has to be a function of two variables. So as long as you stay with $F \left( \begin{smallmatrix} a_1 & a_2 & \cdots & a_k \\ b_1 & b_2 & \cdots & b_{\ell} \end{smallmatrix} \mid z\right)$ and connect this together with field operations, I think Abhyankar is telling you you can never express the roots of a general sextic.
Both of my comments address the question of expressing the roots of $x_n t^n + \cdots + x_1 t + x_0$ as functions of $(x_n, \ldots, x_1, x_0)$ by uniform formulas. I am not addressing the question of whether any particular algebraic number might happen to be equal to a hypergeometric function. I see that you already asked a very hard question along those lines about expressing algebraic numbers in terms of exponentials; I expect that proving anything about hypergeometric functions can only be harder.
