Joel has given one interpretation of this question; let me give another one.
I'll begin by recapitulating an observation made above, but I'll phrase it differently: I claim that viewing the nonstandard number as providing power via its size is misguided. In particular, you state that a machine equipped with a nonstandard number $\alpha$ can solve the Halting Problem by asking, "Does $\Phi_e(e)$ halt by time $\alpha$?" However, nonstandard models of PA can believe false $\Sigma^0_1$ facts - maybe $\Phi_e(e)$ doesn't actually halt, but our model thinks it halts at some nonstandard stage $\beta$.
For example, let $\Phi_e$ be the Turing machine that ignores its input and just searches for a contradiction in PA. Of course $\Phi_e(e)$ doesn't halt; but supposing $M\models$ PA + $\neg$Con(PA), $M$ will think that it halts, and so your idea of a computation will get the halting problem wrong. (More generally, the informal idea you've described will merely let you compute a possibly proper superset of the halting problem.)
So I think the "power-through-size" idea is dangerous (and for instance this is embodied in Joel's observation that there are partial computable functions that are not partial computable with nonstandard help.
Instead, I want to look at power-through-structure. The idea here is that we can think of standard natural numbers as finite objects with a lot of properties (e.g. their prime factorizations). A nonstandard natural number carries the same kinds of information, but much more of it. In particular:
Let $M$ be a possibly nonstandard model of PA. For $n\in M$, let $string(n)$ be the infinite binary string whose $k$th bit is $1$ iff $M\models p_k\vert n$, where $p_k$ is the (numeral representing the) $k$th prime.
(Here by "infinite string," I mean infinite string in the usual sense: it has only standard bits. In particular, by overspill $M$ won't "see" any infinite strings in this sense if $M$ is nonstandard.)
Now let the standard system of $M$, $SS(M)$, be $\{string(n): n\in M\}$. If $M$ is standard, this is boring: it consists exactly of those strings with finitely many $1$s. It turns out, though, that if $M$ is nonstandard, the standard system is much more interesting:
If $M$ is a nonstandard model of PA (in fact, much less), then $SS(M)$ is a Scott set: it is closed under Turing reducibility and joins, and if $T$ is an infinite binary tree in $SS(M)$, then there is an infinite path through $T$ in $M$.
In fact, this is an exact characterization in the countable case:
Let $A$ be a countable set of infinite binary strings. Then the following are equivalent: (i) $A$ is a Scott set. (ii) $A=SS(M)$ for some countable nonstandard model $M$ of PA.
This is true also if we replace "countable" by "of size $\le \aleph_1$," and it is a major open question if the size condition can be completely removed.