Surreal numbers, ultrapowers of $\Bbb R$, ordinal-valued functions and the slow-growing hierarchy

This kind of analysis is very well understood in ultrapowers, and one often sees this kind of thinking with ultrapowers, where one performs calculations with the representing function for an object. What you call $\omega$ is in fact just a particular nonstandard natural number in $\mathbb{N}^*$, and one may perform the same kind of calculations with any nonstandard number. That is, if you had used some other sequence than $(1,2,3,\ldots)$, then you could have done the same kind of calculations.

To probe a bit further, one might ask: is your $\omega$ prime in the nonstandard natural numbers $\mathbb{N}^*$? Well, the answer depends on the ultrafilter $\mu$ that you use in your ultrapower. If $\mu$ happens to concentrate on the set of prime numbers, then it follows by the Los theorem that $\omega$ will be a prime non-standard number. But if $\mu$ concentrates on the composite numbers, then $\omega$ will be composite. More generally, $\omega$ has all and only the properties that hold on a $\mu$-large set, for if $\mu$-almost every $n$ has property $p$, expressible in the language with respect to which you took the ultrapower, then by Los it follows that $\omega$ also has property $p$, and conversely.

So the properties of your element $\omega$ are tightly connected with the particular ultrafilter $\mu$ that you use, and indeed, those properties determine $\mu$, since every set of natural numbers $A\subset\mathbb{N}$ can be regarded as a property and the situation will be that $\omega\in A^*$ just in case $A\in\mu$. This is the sense in which $\omega$ has property $A$ just in case $\mu$ concentrates on $A$.

Actually, there is a strong sense in which your construction has not really determined a specific value for $\omega$. Consider any ultrapower $\mathbb{N}^*=\mathbb{N}^{\mathbb{N}}/\nu$ of $\mathbb{N}$ by any ultrafilter $\nu$, and let $N$ be any nonstandard natural number in this version of $\mathbb{N}^*$. Then I claim that there is another ultrafilter $\mu$, whose ultrapower $\mathbb{N}^{\mathbb{N}}/\mu$ maps elementarily into the original $\mathbb{N}^*$, but where the $\omega$ as determined by $\mu$ maps exactly to $N$, the original arbitrary nonstandard number $N$. To build $\mu$, observe simply that $N=[f]_\nu$ for some function $f:\mathbb{N}\to\mathbb{N}$, and let $\mu=f*\nu$, so that $X\in\mu\iff f^{-1}(X)\in\nu$. It follows that $[g]_\mu\mapsto [g\circ f]_\nu$ is an elementary embedding of the ultrapower by $\mu$ into the ultrapower by $\nu$, and furthermore since $[id]_\mu\mapsto [f]_\nu$, we see that the $\omega$ defined with respect to $\mu$ maps to $N$, as I claimed. In this sense, what you call $\omega$ is simply an arbitrary nonstandard natural number, whose properties are left to be determined by the particular ultrafilter that you use to form the ultrapower.

Alec, I'm afraid your construction does not work. You claim you obtain the surreals by forming the Dedekindean Completion of what you call the surrationals. However, since No--the surreals-- is not Dedekindean Complete, your claim is false. In fact, the Dedekindean Completion of No has "cardinality" $2^{\aleph_{On}}$, which is not even well-defined in NBG (with Global Choice).

No is a (fully) saturated model for the theory of real-closed ordered fields and it has long been known that no such model is Dedekindean complete. On the other hand, there are real-closed fields that are $\kappa$-saturated but not (fully) saturated that are Dedekindean Complete. See, for example, H. J. Keisler and J. Schmerl Making the hyperreal line both saturated and complete. J. Symbolic Logic 56 (1991), no. 3, 1016–1025.

(beginning of post deleted due to a missed error, thanks to Philip Ehrlich)

It is possible to extend the usual hyperoperation sequence $\mathcal{H}_\omega$ on $\mathbb{N}$ to a recursive sequence of operations $\mathcal{H}$ on $O_n$, with the following result:

$$\mathcal{H}_{_\Omega}(\alpha,\beta)=\begin{cases} \mathcal{S}\alpha, & \text{if} \ \Omega=0. \\ \alpha, & \text{if} \ \Omega=1 \ \text{and} \ \beta=0. \\ \mathcal{S}\alpha,& \text{if} \ \Omega=1 \ \text{and} \ \beta=1. \\ 0, & \text{if} \ \Omega=2 \ \text{and} \ \beta=0. \\ \alpha, & \text{if} \ \Omega=2 \ \text{and} \ \beta=1. \\ 1, & \text{if} \ \Omega\geq3 \ \text{and} \ \beta=0. \\ \alpha, & \text{if} \ \Omega\geq3 \ \text{and} \ \beta=1. \\ \mathcal{H}_{_{\Omega-1}}\big(\alpha,\mathcal{H}_{_\Omega}(\alpha,\beta-1)\big), & \text{if} \ \Omega=\mathcal{S}\bigcup\Omega \ \text{and} \ 1<\beta=\mathcal{S}\bigcup\beta. \\ \bigcup_{\delta<\beta}\mathcal{H}_{_\Omega}(\alpha,\delta), & \text{if} \ \Omega=\mathcal{S}\bigcup\Omega \ \text{and} \ \ 1<\beta=\bigcup\beta . \\ \bigcup_{\rho<\Omega}\mathcal{H}_{_\rho}(\alpha,\beta), & \text{if} \ 0\neq\Omega=\bigcup\Omega \ \text{and} \ 1<\beta=\mathcal{S}\bigcup\beta. \\ \bigcup_{\rho<\Omega}\bigcup_{\delta<\beta}\mathcal{H}_{_\rho}(\alpha,\delta), & \text{if} \ 0\neq\Omega=\bigcup\Omega \ \text{and} \ 1<\beta=\bigcup\beta. \\ \end{cases}$$

This 'transfinite hyperoperation sequence' is a very 'large' object in the sense that each hyperoperation in the sequence is a proper class, but it is well defined under the appropriate reflection principle or under ETR and satisfies nice relations like $\mathcal{H}_4(\omega,\omega)=^\omega\omega=\omega^{\omega^{\omega^{\dots}}}$ in your question, and $\mathcal{H}_5(\omega,\omega)=^{^{^{\dots}\omega}\omega}\omega$, so on and so forth. I hope this assists with understanding what is happening with the hyperoperation-related piece of the questions you've mentioned, as I find them fascinating as well!