For an upper bound, you have $E\min(X,Y)\le \min(E X,E Y)$, and so $$ E\min(c^TX_1,c^TX_2)\le c^T\cdot(1/2,\ldots,1/2)=\frac12\sum_{i=1}^n c_i.$$ For the lower bound, notice first that $\min(a,b)=(a+b-|a-b|)/2$, whence $$ E\min(c^TX_1,c^TX_2) = \frac12\sum_{i=1}^n c_i -\frac12E|c^TX_1-c^TX_2|. $$
It remains to upper-bound the latter term. Hölder's inequality comes to mind: we can bound $|c^T(X_1-X_2)|$ by $||c||_2||X_1-X_2||_2$, or by $ ||c||_1||X_1-X_2||_\infty $, or, say, by $ ||c||_\infty||X_1-X_2||_1 $.
Let's see where the first bound leads. We have $$E||X_1-X_2||_2^2=\sum_{i=1}^nE(X_1(i)-X_2(i))^2 =\frac{1}{6}n, $$ the latter is a routine calculation, see https://en.wikipedia.org/wiki/Triangular_distribution . Now $E||X_1-X_2||_2 = E\sqrt{||X_1-X_2||_2^2||} \le\sqrt{E||X_1-X_2||_2^2}=\sqrt{n/6} $.
This yields a lower bound of $$ \frac12\sum_{i=1}^n c_i -\frac{||c||_2}2\sqrt{\frac{n}6} \le E\min(c^TX_1,c^TX_2). $$
You'll get other estimates via the other applications of Hölder, which will be better or worse depending on $||c||_p$, for $p\in[1,\infty]$.