show that $L(\ell_1,X) \cong \ell_\infty(X)$ via $\phi(T)(i) = T(e_i)$ holds isometrically I'd like to get a proof verification.
Suppose $\phi(T) = \phi(T')$. Then they are the same as elements in $\ell_\infty(X)$ (i.e., $\phi(T)(i)= \phi(T')(i)$ for all $i$). But then $T(e_i) = T'(e_i)$ by the definition of $\phi$. Since $T,T'$ are linear and have the same behavior on basis vectors $e_i$, $T=T'$ and $\phi$ is one-to-one. Fix $x \in \ell_\infty(X)$. Define a linear map $T$ given by $T(e_i) =x_i$ for all $i$. Then $T \in L(\ell_1,X)$ and $\phi(T)(i) = T(e_i) = x_i$ for all $i$, giving that $\phi(T) = x$. So $\phi$ is surjective and thus bijective. But for any $a,b$, $\phi(aT +bT')(i) = (aT+bT')(e_i) = aT(e_i)+bT'(e_i) = a\phi(T)(i)+ b\phi(T')(i) = (a\phi(T)+b\phi(T'))(i).$
So $\phi(aT +bT') = a\phi(T)+b\phi(T')$, $\phi$ is an isomorphism and $L(\ell_1,X) \cong  \ell_\infty(X)$.
Fix $T \in L(\ell_1,X)$. Then
\begin{align*}
 \|\phi(T)\|_{\ell_\infty(X)}
 &= \sup_i \|\phi(T)(i)\|_{X} \\
 &= \sup_i \|T(e_i)\|_{X} \\
 &= \sup_{\|x\|_{\ell_1(X)} \leq 1} \|T(x)\|_{X}\\
 &= \|T\|_{L(\ell_1,X)},
 \end{align*}
where the third line follows from that $T$ is linear $\implies$ $\|Tx\|_X$ is maximized for some basis vector $e_i$. So $\phi$ is an isometry.
 A: Injectivity: your argument is kind of fine, but you don't use that $T$ is bounded. And you need it to pass from "equal on basis elements" to "equal everywhere".
Surjectivity: you don't say that you extend by linearity. You don't show that $T$ is bounded. And you need this to know that you can extend from a dense subset to the whole thing.
Linearity: it's fine.
Isometry: "is maximized for some basis vector $e_i$". No. Not only is the norm not in general maximized on a basis element (even in finite dimension); it is often not a maximum.
As a general comment, your arguments never use that $\ell_1$ is $\ell_1$; that should tell you that you are glossing over something.

What happens here is that if $a\in\ell_1$ and $T$ is bounded, then
$$
Ta=\sum_ja_jTe_j.
$$
Using the continuity of the norm,
$$
\|Ta\|\leq\sum_j|a_j|\,\|Te_j\|\leq\sup\{\|Te_j\|:\ j\}\,\sum_j|a_j|=\|(Te_j)_j\|_\infty\,\|a\|_1=\|\phi(T)\|_\infty\,\|a\|_1.
$$
So $\|T\|\leq\|\phi(T)\|_\infty$. Given $\varepsilon>0$, there exists $j$ such that $\|\phi(T)\|_\infty<\|Te_j\|+\varepsilon$. As $\|e_j\|=1$,
$$
\|T\|\geq\|Te_j\|>\|\phi(T)\|_\infty-\varepsilon. 
$$
As this can be done for any $\varepsilon>0$, this shows the reverse inequality and then $\|T\|=\|\phi(T)\|_\infty$, which shows that $\phi$ is an isometry.
