Robin Chapman's answer is very apropos. Here is a theoretical answer that points out a subtlety in the question.
First, recall that the primitive recursive functions are the smallest class of functions on $\mathbb{N}$ that:
- Includes the constant zero function, the successor function, and all projection functions;
- is closed under composition;
- and is closed under primitive recursion.
Let's call this class of functions $\operatorname{PR}(\emptyset)$. For any set $A$ of number theoretic functions, we can define a more general class $\operatorname{PR}(A)$ as the smallest class of functions that satisfies the above properties and also includes every function in $A$.
If every function in $A$ is computable, then every function in $\operatorname{PR}(A)$ is computable. Moreover, if every function in $A$ is total then every function in $\operatorname{PR}(A)$ is total.
It would be trivial, assuming $A$ is finite (or, more generally just explicitly enumerated), to create a programming language such that every program in the language computes a function in $\operatorname{PA}(A)$ and every such function has a program in the language. The language simply has primitives for all the functions in $A$ and for the basic primitive recursive functions, along with operators for composition and primitive recursion.
Therefore, one answer to "I'm curious about how "much" we can compute with a formalism that guarantees halting." is "For any total computable function there is such a formalism" and more generally this is true for any effective sequence of total computable functions.
The main thing that such a system cannot have is a universal function, provided the system has some basic closure properties.
Addendum
Since several people pointed out the same thing, I may as well add some value to the answer by including the standard proof of my remark about universal functions. A universal two-place function, for some system, is a function $g(i,k)$ such that for every one-place function $f(k)$ in the system there is some natural number $e$ with $\lambda k .f(k) = \lambda k. g(e,k)$. Suppose that $g$ is such a function; define a function $h$ as $h(k) = g(k,k) + 1$. Then $h$ has some index $e$. Thus $g(e,e) = h(e)$ by the definition of $g$ and $e$, and $h(e) = g(e,e) + 1$ by construction of $h$. This is impossible, so any system of total functions that allows me to form the functions like $h$ cannot have a universal function $g$.
In particular, for any class of functions $A$ and any function $g(j,i)$ in $\operatorname{PA}(A)$, the function $h(n) = g(n,n)+1$ is in $\operatorname{PR}(A)$. So this class will not have a universal function provided that every function in $A$ is total. Of course if you let $A$ be the set of all partial computable functions then $\operatorname{PR}(A)$ is also the set of all partial computable functions and so it does contain a universal function (which is not total).
From this point of view, the limitation is not on whether particular computable functions can be included; the limitation is on the internal structure of any particular effective class of functions.