For the asymptotic $\ell_1$ space, I believe we get $\|\sum_{i=1}^{2n} e_i\|=\max \{1,n/2\}$ and $\|\sum_{i=1}^{2n+1} e_i\|=\max\{1,(n+1)/2\}$, which is in line with the answer by @SArgyros. For the asymptotic $c_0$ space, I think the answer is much more complicated. It probably involves the Ackermann functions.

It says that $f(x,y) \geq f(z,w)$ if and only if $x\geq z$ and $y\geq w$. Choose $x>z$ and $y<w$. Then the condition $(x\geq z)\wedge (y\geq w)$ is not satisfied, and therefore the condition $f(x,y)\geqslant f(z,w)$ is not satisfied. Similarly, since $z<x$, the condition $(z\geq x)\wedge (w\geq y)$ is not satisfied, so the condition $f(z,w) \geq f(x,y)$ is not satisfied. Therefore $f(x,y)<f(z,w)$ and $f(z,w)<f(x,y)$.

Is this in reference to my original comment, which is now deleted?

How is condition 4 possible?

Are you asking this: If $\varphi:L_p(\mu,X)^*\to L_q(\mu,X^*)$ is an isometric isomorphism, $N\in \Sigma$, and $K\in L_p(\mu,X)^*$ is such that $K(f1_N)=0$ for all $f\in L_p(\mu,X)$, then $\varphi(K)=0$ a.e. on $N$? The answer to this would be no. Consider $\Omega=[0,1]$, $\mu$ Lebesgue measure. Let $\psi:L_p^*\to L_q$ be the usual isometric isomorphism and let $\varphi K (x) = \psi K(1-x)$. Then any function which vanishes on $[0,1/2]$ would satisfy $K(f 1_{[1/2,1]})=0$ for all $f\in L_p$.

By "the Tsirelson space", do you mean Tsirelson's original space (which is asymptotic $c_0$) or the Figiel-Johnson Tsirelson space (which is asymptotic $\ell_1$)?

My answer gets you to the conclusion that you want, but it doesn't really explain why. To see why the result is true, you will want to look at the proof that the inclusion of $L_q(\mu,X^*)$ into $L_p(\mu,X)^*$ is an isometric embedding. The fact that it's norm at most $1$ follows from H\:{o}lder. To see why it's isometric, I'd recommend looking first at simple functions.