Ah yes, a fave topic of mine. Basically, there is no universally-agreed on way to do this. The problem is, that, in general, there isn't a unique way to interpolate the values of tetration at integer "height" (which is what the "number of exponents in the 'tower'" may be called). So in theory, you could define it to be anything.
In the case of exponentiation, one has the useful identity $a^{n + m} = a^n a^m$, which enables for a "natural" extension to non-integer values of the exponent. Namely, you can see, for example, that $a^1 = a^{1/2 + 1/2} = (a^{1/2})^2$, from which we can say that we need to define $a^{1/2} = \sqrt{a}$ if we want that identity to hold in the extended exponentiation. No such identities exist for tetration.
You may also want to look at Qiaochu Yuan's answer here, where he explores some of this from a viewpoint of higher math:
https://math.stackexchange.com/a/56710/11172
One could, perhaps, compare this problem to the question of the interpolation of factorial $n!$ to non-integer values of $n$. There is, in general, no simple identity that provides a natural extension for this, either. But, when an extension is desired, the usual choice is to use what is called the "Gamma function", defined by
$$\Gamma(x) = \int_{0}^{\infty} e^{-t} t^{x-1} dt$$.
Then, you can extend $n!$ to non-integer $x$ by $x! = \Gamma(x+1)$. However, usually one does not use $x!$ for non-integer factorials, but rather the Gamma function notation.
One can give a uniqueness theorem involving soem simple analytical conditions; it is called the Bohr-Mullerup theorem. In addition, the gamma function has various nice number-theoretic and analytic properties, and turns up in a number of different areas of math.
But in the case of tetration, there are no nice integral representations known. Henryk Trappmann and some others recently proved a theorem that gives a simple uniqueness criterion for the inverse of tetration (with respect to the "height") here, presuming extension not just to the real, but the complex numbers:
http://www.ils.uec.ac.jp/~dima/PAPERS/2009uniabel.pdf
The solution that satisfies the condition is one that was developed by Hellmuth Kneser in the 1940s. I call it "Kneser's tetrational function" or simply "Kneser's function". It defies simple description.
On this site:
http://math.eretrandre.org/tetrationforum/index.php
an algorithm was posted to compute the Kneser solution (though I'm not sure if it's been proven) for various bases of tetration. Using this solution, the answer to your question would be
$$^{4/3} 3_\mathrm{Kneser} = 4.834730793026332...$$
Other interpolations for tetration have been proposed, some of which give different results. But this is the only one that seems to satisfy "nice" properties like analyticity and has a simple uniqueness theorem via its inverse. Yet as I said in the beginning, I don't believe that it's universally agreed by the general mathematical community that this is "the" answer.
It seems to me that before much more progress can be made in the calculus of ${}^xy$, more fundamental questions have to be answereed, such as, how to define ${}^xy$ for rational $x$? It's clear how the OP's definition works if $x$ is a non-negative integer; but how do we define ${}^xy$ if, say, $x = 7/2$? What then is "one-half" of an occurrance of $x$ in the exponential "tower" which is supposed to be ${}^xy$?
I am reminded here of the way $x^y$ is extended from integers through the reals, by starting with a careful, consistent and believable definition of $(p / q)^{(r / s)}$ for integral $p, q, r, s$; once we have that, a simple, consistent and believable continuity argument allows us to accept a definition of $x^y$ for real $x, y > 0$. We know what $(p / q)^r = (p^r / q^r)$ means; we know what it means for a positive real $z$ to satisfy $z^s = (p / q)^r$, so we can get a handle on $(p / q)^{(r / s)}$ from which, by continuity, we can generalize to $x^y$. I think an analogous method is needed here, but I don't know what it is. But I think my question of the preceding paragraph might be worth considering early on in this game.
Of course, perhaps there is a (reasonably) simple, consistent and believable argument to contruct ${}^xy$ using $\exp()$, $\log()$, etc., or some sort of differential or similar equation ${}^xy$ must satisfy, or perhaps one could learn something from the $\Gamma$ function and factorials here which would bypass, at least temporarily, the need to address how ${}^{(p / q)}(r / s)$ is supposed to work, but sooner or later the question will have to be faced, I'll warrant.
This is an interesting, though speculative, arena and I am glad to have participated. But until I can answer my own questions to my better satisfaction, I will refrain from further
remarks, except to bid those who are ready to climb such unknown heights, "Excelsior!
Hope this helps, at least with the spirit of the adventure if not with the direction. Happy New Year,
and as always,
Fiat Lux!!!
Best Answer
As Will Jagy already pointed out, there is no accepted solution for this. There is a formal procedere which can sometimes lead to a meaningful/approximate answer; but this is then dependent on the convergence of some series, which occur in that procedere, and also, in which way we want to make sense of noninteger powers of negative or complex numbers.
I mean the procedere which applies the Schröder-function on a recentered powerseries.
The formal way to do this is to
I've just tried this with the base $b=i$ in Pari/GP (that gave also $ \small t \sim 0.438282936727 + 0.360592471871 i $ and $ \small u \sim -0.566417330285 + 0.688453227108 i$) and for small integer heights h this gives good approximations (although we have complex powerseries involved(!)). However, we see in the last step, that the log $u$ of the fixpoint has to be taken to the h'th power - and this means for your question a complex value to the $i$'th power. This is not unique and the powers of this ("incidentally" selected) value occur then in the formula for the inverse Schröder-function.
Now, after I simply inserted $h=i$ and let Pari/GP determine the result using $\sigma u^i $ for the $i$'th tetrate (beginning from $x_0=1$) with basis $i$ then I arrive at something like $x_i = i^{\text{^^}i} \sim 0.500129061734+0.324266941213 i $ .
This whole procedure has - even for the case of a real base and real fixpoints, the further unsolved disadvantage, that the result will be depending on the selection of the fixpoint $t$ . Moreover, even for simple fractional $h$ , say $h=0.5$ , the occuring series are not or only roughly converging correctly, such that the half-iterate from $x$ to $x_{0.5}$ and then the further half-iterate from $x_{0.5}$ to $x_1$ is only approximate with (practically necessarily) truncated power series. So besides the non-uniqueness at the point, where we raise the complex $u$ to the $i$'th power in our specific example, there is also basically a problem of convergence with that whole procedure.
Note, there is one claim that a general solution for the tetration to complex heights was found; look at citizendium at the article of Dmitri Kouznetsov, but I've not yet seen that it had externally been confirmed (and I cannot do it myself). Also we have in the tetration-forum the claim, that we can uniquely determine fractional tetrates at least for some "nontrivial" bases, which is basically derived from Kneser's ansatz for the fractional iteration with base $b=e$. Look at the posts of user "Sheldonison" (Levenstein) who even provides a set of Pari/GP-procedures to compute fractional tetrates for a range of bases outside the "easy" range $1 \ldots e^{1/e} $ of real bases. Unfortunately I'm not yet capable to judge over that claims of Kouznetsov and of Levenstein.
[update] There has been an attempt today to insert in my answer a statement, that the Kouznetsov-method were "the official" tetration, based on some consideration of reasons. I reject that. Without more qualified echo in the professional mathematical community it is not my part to claim such and generate avoidable obfuscation. Please do not attempt to extend the focus of my answer in such a way - you can always put it in another answer.