Alas, no such inequality can hold. Suppose that the symmetric $X_i$ take values $\pm 1$ and $t=n-1$. Then 
$$
  \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] =(n+1)2^{1-n}
$$
but
$$\Pr[|\sum_{i \in [n]} X_i| \ge t]=2^{1-n} \,.
$$