$\newcommand\dotcup{\mathbin{\dot\cup}}$I shared your confusion until a year ago, when Markus Haase explained to me that \eqref{2} can actually be rewritten as assertion (ii) in the following theorem, which I find much clearer.
Theorem. Let E be a Banach space, let $(\Omega,\mu)$ be a measure space, let $p \in [1,\infty)$ and let $I$ be a non-empty index set. For each $i \in I$ let $T_i: E \to L^p := L^p(\Omega,\mu)$ be a bounded linear operator. The following are equivalent:
(i) The operator family $(T_i)_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $f \in E$ there exists $0 \le h \in L^p$ of norm $\lVert h\rVert_{L^p} \le C \lVert f\rVert_{E}$ such that $\lvert T_if\rvert \le h$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).
(ii) There exists a constant $C$ with the following property:
for every finite measurable partition $\Omega = A_1 \dotcup \dotsb \dotcup A_n$, all indices $i_1, \dotsc, i_n \in I$, and every $f \in E$ one has $\Bigl\lVert \sum_{k=1}^n 1_{A_k} T_{i_k} f \Bigr\rVert_{L^p} \le C \lVert f\rVert_{E}$.
Proof.
"(i) $\Rightarrow$ (ii)"
This implication is straighforward to check.
"(ii) $\Rightarrow$ (i)"
Let $\mathcal{F}$ denote the set of all non-empty finite subsets of $I$, which is directed with respect to set inclusion.
Fix $f \in E$.
For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} \lvert T_if\rvert \in L^p$ (where $\lvert\cdot\rvert$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\lVert f\rVert_{E}$. To see this, enumerate the elements of $F$ as $i_1, \dotsc, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dotsc, A_n$ such that
$$
\bigvee_{i \in F} \lvert T_if\rvert
=
\sum_{k=1}^n 1_{A_k} \lvert T_{i_k} f\rvert
=
\Bigl\lVert \sum_{k=1}^n 1_{A_k} T_{i_k} f \Bigr\rVert
.
$$
Since the net $\bigl(\bigvee_{i \in F} |T_if|\bigr)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\lVert f\rVert_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\lVert f\rVert_{E}$ (here was used that $p \in [1,\infty)$).
Clearly, $h$ satisfies the property claimed in (i). $\quad \square$
Remarks.
(a) Assertion (i) is a convenient way to write down a maximal inequality if the index set $I$ is not assumed to be countable and if one does not make any regularity assumptions regarding the dependence of $T_i$ on $i$. Writing down the maximal operator before knowing whether a maximal inequality holds leads to measurability problems in this case. (But once the maximal inequality is established, one can define the maximal operator by using the supremum within the ordered space $L^p$ rather than the almost everywhere supremum.)
In cases where the maximal operator can be reasonably defined as a pointwise supremum assertion (i) is equivalent to the $L^p$-boundedness of the maximal operator.
(b) In the situation of the question, assertion (ii) of the theorem is precisely the estimate \eqref{2} for simple functions $t: \mathbb{R}^n \to (0,1)$. So this shows, in particular, that it suffices to consider simple functions in \eqref{2}.
This does not only resolve all measurability issues regarding $t$; rather, in the situation of the theorem such a measurability is not even defined since the index set $I$ is not assumed to carry a $\sigma$-algebra.
In the same vein, note that assertion (ii) makes sense in this general setting, while for an object such as $T_{i(x)}f(x)$ for a non-simple function $i: \Omega \to I$ it is not even clear how to define it if the $T_i$ do not have any particular structure.
(c) The measure space $(\Omega, \mu)$ is not assumed to be $\sigma$-finite. In fact, the proof shows that the theorem does not really rely on the structure of $L^p$-spaces, but stays true on Banach lattices where every increasing norm bounded net is norm convergent (these are the so-called KB-spaces, of which $L^p$ for $p \in [1,\infty)$ are examples).
To make sense of (ii) one needs to replace the multiplication with indicator functions with band projections in this abstract setting.
(d) The proof shows that $C$ can be chosen as the same number in (i) and (ii).
(e) Fun fact: The existence of $C$ is actually redundant in (i).
Indeed, if one only assumes that for each $f \in E$ there exists $0 \le h \in L^p$ such that $\lvert T_i f\rvert \le h$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $h$ can always be chosen to satisfy $\lVert h\rVert_{L^p} \le C \lVert f\rVert_{E}$.
This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in the preprint Order boundedness and order continuity properties of positive operator semigroups with Michael Kaplin.
Best Answer
I think it is not possible to do much better than $\frac{1}{\sqrt n}$. More precisely, I believe the best $\alpha$ is $\frac{1}{\sqrt n}$ whenever $n$ is a power of $2$, and therefore (since $\alpha$ is non-decreasing in $n$) at most $\frac{\sqrt{2}}{\sqrt n}$ in general.
Indeed, consider $f(x)=\frac 1 2 \|A^{-1} x\|_2^2$ for a linear invertible map $A$ on $\mathbf{R}^n$. If I understand correctly the definition of $1$-strongly convex function, $f$ is $1$-strongly convex iff the $2\to 2$ norm of $A$ is $\leq 1$. For this specific class of examples, your question reduces to "is there a constant independent of $n$ such that $\alpha \|A\|_{\infty \to \infty} \leq \|A\|_{2\to 2}$ for every linear map $A$ on $\mathbf{R}^n$". This is clearly false, for example for a Hadamard matrix: its $2 \to 2$ norm is $\sqrt{n}$ and $\infty\to \infty$ norm is $n$. The claim in the first paragraph of the answer therefore follows from the existence of Hadamard matrices of size $n$ whenever $n$ is a power of $2$.