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 $p\in[1,\infty]$.