The Peano's square-filling curve  $p:I\to I^2$ turn's out to be Hölder continuous with exponent $1/2$ on the unit interval $I$ (a quick way to see it, is to note that $p$ is a fixed point of a suitable contraction $T:C(I,I^2)\to C(I,I^2)$, and the non-empty, closed subset of curves with modulus of continuity $\omega(t):=ct^{1/2}$ is  $T$-invariant, for a suitable choice  of $c$, so that $p$ is therein). 

For the same reason, the more general analogous $n$-cube-filling curves $I\to I^n$ (e.g. described in the same Peano paper) are Hölder continuous with exponent $1/n$.

On the other hand, for any $1\le k\le n$, by elementary considerations on  Hausdorff measures, no $\alpha$-Hölder continuous  map $I^k\to I^n$ with exponent $\alpha > k/n $ can be surjective. The natural questions are therefore: 

>Given $1\le k\le n$, does there exist an $\alpha$-Hölder continuous map $I^k\to I^n$ with exponent $\alpha=k/n$?  Otherwise, what is the best   exponent $\alpha$ obtainable for such a surjective map?  In particular, is there a simple construction for the case $I^2\to I^3$? (Actually, we may focus on this  last question, which appears to be the simpler non-trivial case).

Summing up the above remarks, the answer is affirmative if $k=1$ or if $k=n$, and we may also note that if for a pair $(k,n)$ there is such a surjective map $q:I^k\to I^n$, then the map $(x_1,\dots,x_m)\mapsto (q(x_1), q(x_2),\dots ,q(x_m))$ is also a surjective map $I^{mk}\to I^{mn}$ with the same exponent $k/n$ of $q$. Also, we may consider compositions of maps, so that affirmative answers for $(k,n)$ and $(n,m)$ imply the affirmative answer for $(k,m)$.