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
- determine a fixpoint t for the exponentiation with base b such that $b^t = t$ (here our base is $b=i$ ). Also denote the log of $t$ as $\log(t)=u$
- define the helper function $f(x) = t^x-1 $
- compute the (lower) triangular Carleman-/Bell-matrix for $f(x)$ , call it for instance $U$
- diagonalize the matrix $U$ into $M$, $D$, $W (=M^{-1})$ . Then in the diagonal of $D$ are the consecutive powers of $u$
- define the Schröder-function $s$ from the second column of $M$ such that
$ s(x) = \sum_{k=1}^\infty m_{k,1} x^k $ and its inverse from the second column of $W$ such that $ s^{-1}(x) = \sum_{k=1}^\infty w_{k,1} x^k $
- compute the schröder-value $\sigma = s(x/t-1) $ for some argument $x$. (For instance $x=1$ -this is also tacitly assumed if we write b^^h for the h'th tetrate with base b)
- compute the function value for the h'th tetration-height by $x_h = (s^{-1}(\sigma u^h)+1) \cdot t$
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.
Late update: here is a picture showing fractional iteration based on the Schroeder/Koenigs-solution for the tetration.
The start is taken for $z_0=1$ and from this it is tetrated with real heights from $z_0$ to $z_7$ giving the nearly spiral trajectory towards the (first) fixpoint $z_\infty$.
Iteration with (purely) imaginary heights follow the red line direktly from $z_0$ to $z_\infty$ The point which is arrived at after iteration "$î$ times" is marked as $z_î$ (near the fixpoint at the end of the red line)
[author's reservation] 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.
