The problem asked (without independence) can be solved.
Fix nonnegative real numbers $a_1, …, a_n$. Define $M = \sum_{i=1}^n a_i$ and assume $M \geq 1$. Define $$\mathcal{X} = \{0, a_1\} \times \{0, a_2\} \times … \times \{0, a_n\} $$ We solve the problem by first solving a simpler problem ("Problem 1" below) that fixes a specific value for $E[X]$.
Problem 1: Fix $c \in [1,M]$. We want to find a random vector $(X_1, ..., X_n)$ to solve \begin{align} \mbox{Minimize:} \quad & P\left[\sum_{i=1}^n X_i \geq c\right] \\ \mbox{Subject to:} \quad & E\left[\sum_{i=1}^n X_i\right] = c \\ & (X_1, ..., X_n) \in \mathcal{X} \end{align}
To solve problem 1, define $$ g_c = \max_{(x_1, ..., x_n) \in \mathcal{X}} \left\{\sum_{i=1}^n x_i : \sum_{i=1}^n x_i < c\right\} $$ That is, $g_c$ is the largest value of $\sum_{i=1}^n x_i$ over all vectors $(x_1, ..., x_n) \in \mathcal{X}$ with components that sum to less than $c$. The max is achievable because $\mathcal{X}$ is a finite set that contains the all-zero vector. Note that $0 \leq g_c < c \leq M$. Define $\vec{x}_c$ as an element of $\mathcal{X}$ with a sum of components that is equal to $g_c$. Define $$p_c = \frac{c-g_c}{M-g_c}$$ Observe that $0< p_c \leq 1$. Define the random vector $$ (Z_1, ..., Z_n) = \left\{ \begin{array}{ll} (a_1, ..., a_n) &\mbox{ with prob $p_c$} \\ \vec{x}_c & \mbox{ with prob $1-p_c$} \end{array} \right.$$ It is not difficult to show that $E[\sum_{i=1}^nZ_i] = c$ and $$P\left[\sum_{i=1}^n Z_i \geq c\right] = p_c $$
Claim
The random vector $(Z_1, ..., Z_n)$ solves problem 1. In particular, the minimum probability is: $$ P\left[\sum_{i=1}^n Z_i \geq c\right] = p_c = \frac{c-g_c}{M-g_c}$$
Proof
Define $\lambda_c = \frac{1}{M-g_c}$ and note that $\lambda_c>0$. Define $$ A = \left\{\left(E[X_1], ..., E[X_n], P\left[\sum_{i=1}^n X_i \geq c\right]\right) \in \mathbb{R}^{n+1} : (X_1, ..., X_n) \in \mathcal{X}\right\} $$ Consider the following two problems:
Constrained problem: \begin{align} \mbox{Minimize:} \quad & x_{n+1}\\ \mbox{Subject to:} \quad & \sum_{i=1}^n x_i = c \\ & (x_1, ..., x_{n+1}) \in A \end{align}
Unconstrained problem: \begin{align} \mbox{Minimize:} \quad & x_{n+1} + \lambda_c\left(c-\sum_{i=1}^n x_i\right) \\ \mbox{Subject to:} \quad & (x_1, ..., x_{n+1}) \in A \end{align}
Observe that the constrained problem is equivalent to Problem 1. Also observe by standard Lagrange multiplier theory that if $\vec{x}^*$ is a solution to the unconstrained problem that satisfies $\sum_{i=1}^n x_i^*=c$, then it also solves the constrained problem. Indeed, such a vector $\vec{x}^*$ would trivially satisfy all constraints of the constrained problem, and for any other vector $\vec{w}=(w_1, ..., w_n, w_{n+1})$ that satisfies the constraints of the constrained problem, we have (since $\vec{w} \in A$): $$ x_{n+1}^* + \lambda_c\underbrace{\left(c-\sum_{i=1}^n x_i^*\right)}_{0} \leq w_{n+1} + \lambda_c\underbrace{\left(c-\sum_{i=1}^n w_i\right)}_{0} $$ and so $x_{n+1}^*\leq w_{n+1}$. Define the function $f:\mathcal{X}\rightarrow\mathbb{R}$ by: $$ f(x_1, ..., x_n) = 1\left\{\sum_{i=1}^n x_i\geq c\right\} + \lambda_c\left(c-\sum_{i=1}^n x_i\right) $$ where $1\{\cdot\}$ denotes an indicator function. It is not difficult to show that $$f(x_1, ..., x_n) \geq p_c \quad \forall (x_1, ..., x_n) \in \mathcal{X}$$ Hence, for all random vectors $(X_1, ..., X_n) \in \mathcal{X}$ we have: $$ f(X_1, ..., X_n) \geq p_c$$ and so for all random vectors $(X_1, ..., X_n) \in \mathcal{X}$ we have $E[f(X_1, ..., X_n)] \geq p_c$, that is: $$ P\left[\sum_{i=1}^n X_i \geq c\right] + \lambda_c (c - E[\sum_{i=1}^n X_i]) \geq p_c $$ However, the particular random vetor $(Z_1, .., Z_n) \in \mathcal{X}$ satisfies $\sum_{i=1}^n Z_i \in \{M, g_c\}$ always (where $0\leq g_c < c\leq M$), and so $$ f(Z_1, ..., Z_n) = p_c \quad , \mbox{ for all realizations of $(Z_1,...,Z_n)$} $$ Hence $$ E[f(Z_1, ..., Z_n)] = p_c $$ Thus, the random vector $(Z_1, ..., Z_n)$ minimizes the expression: $$ P\left[\sum_{i=1}^n X_i \geq c\right] + \lambda_c (c - E[\sum_{i=1}^n X_i]) $$ over all random vectors $(X_1, ... X_n) \in \mathcal{X}$. That is, the random vector $(Z_1, ..., Z_n)$ solves the unconstrained problem. But we already know this random vector satisfies $\sum_{i=1}^n E[Z_i]=c$, hence (by Lagrange multiplier theory) it solves the constrained problem, hence it solves Problem 1. $\Box$
Problem 2: This is the original problem of the question, which does not specify the mean $E[\sum_{i=1}^n X_i] =c$. So we just optimize over $c \in [1, M]$.
$$ p^* = \min_{c \in [1, M]}\left[ \frac{c-g_c}{M-g_c} \right]$$