Informed by Fedor Pakhomov's excellent answer, and using "slow" iterated $Σ_1$-soundness, here is an infinite sequence of sound theories $T$ such that $T_i$ = PA + 1-Con($T_{i+1}$).

Set $T_i$ = PA + "$h'_{g(n)∸i}(n)$ is total"<br/>
using fast-growing or Hardy hierarchy $h$ (either one works), with $h'_α$ being $h_{ε_α}$, and $g(n) = {h'_{ω+1}}^{-1}(n)$, where $f^{-1}(n) = \max(m: ∀m'<m \,\, f(m')<n)$ (or any reasonable variation on this).   Also, $a∸b = \max(a-b,0)$.

Alternatively, almost any reasonable monotonic total recursive $g$ that is sufficiently slow-growing but tending to infinity works, but note that $T$ depends on $g$. The construction also generalizes to other theories extending $Σ^0_1$-PA (the proof uses $Σ^0_1$ induction) by replacing $h'_m(n)$ with a function corresponding to $1+m$ iterations of $Σ_1$-soundness.

Provably in PA, if $g$ is unbounded (equivalently, $h'_{ω+1}$ is total), then $T_i$ (and even PA + "$h'_ω$ is total") is $Σ_1$-sound.  Thus, it suffices to prove that for each $i$, PA + "$g$ is bounded" proves "$h'_{g(n)∸i}(n)$ is total" $⇔$ 1-Con($T_{i+1}$). (However, the quantification over $i$ will be unprovable even in $T_0$).

Working in PA + "$g$ is bounded", let $k$ be the maximum value of $g$. Since $k$ is (externally) nonstandard, $k>i$.  Because 1-Con is unaffected by true $Π^0_1$ statements, in 1-Con(...), we can freely assume/assert that $k$ is the maximum value of $g$.  Thus, 1-Con($T_{i+1}$) $⇔$ 1-Con(PA + "$h'_{k-(i+1)}(n)$ is total") $⇔$ "$h'_{k-i}(n)$ is total"  (the latter follows from standard results in ordinal analysis), as required.