One of the usual ways of proving Tennenbaum's theorem also
applies to many of the theories on your list, and so they
can have no computable nonstandard models.

The proof I have in mind is the following, which I also
explained in [this MO
answer](https://mathoverflow.net/questions/12426/is-there-a-computable-model-of-zfc/12434#12434).
Let $A$ be the set of Turing machine programs that halt on
input $0$ with output $0$, and let $B$ be the set of
programs that halt on input $0$ with output $1$. These sets
are disjoint and [computably
inseparable](http://en.wikipedia.org/wiki/Recursively_inseparable_sets),
meaning that there is no computable $C$ containing $A$ and
disjoint from $B$. Now, suppose that $M$ is a nonstandard
model of arithmetic. Let $d$ be a nonstandard natural
number, and inside $M$, consider the set of programs below
$d$ that halt on input $0$ in at most $d$ steps with output
$0$. In $M$, this is a (nonstandard) finite list, and so
there is a nonstandard number $c$ coding this list of
programs. Now, let $C$ be the set of standard programs $p$
that $M$ thinks appear on the list coded by $c$. This
includes every program in $A$, since all such programs halt
in a standard finite time with output $0$, and hence $M$
will agree that they halt before time $d$. Second, for a
similar reason, this set includes no programs in $B$, since
those programs halt in finite time with output $1$, and $M$
will see that. Finally, the set $C$ is computable from the
operations of $M$, since we need only perform the decoding
procedure to see if a given number $p$ is on the list coded
by $c$. For example, we might use the coding that would
require us merely to check whether $M$ thinks that the
$p^{th}$ binary digit of $c$ is $1$ or not. If the
operations of $M$ were computable, then this would be a
computable procedure, in contradiction to the fact that $A$
and $B$ are computably inseparable. QED

Now, we haven't really used much of PA in this argument.
Any theory $T$ that is able to perform basic Goedel coding
and simulate Turing machine computations will be sufficient
for the argument. This includes any $I\Sigma_n$, even
$I\Sigma_0$, since the operation of a Turing machine is
inductively iterating a very trivial process. So none of
the stronger theories on your list have computable
nonstandard models.

Meanwhile, however, I am unsure about the very weakest
theories on your list, but this argument reduces the
question to: can the given theory prove that for any number
$d$, there is a number $c$ coding the list of Turing
machine programs less than $d$ that halt on input $0$ with
output $0$ in at most $d$ steps?

This is a comparatively simple statement in arithmetic, and
any theory proving it will not have computable nonstandard
models.