I now outline how I would solve it. There likely exists shorter solutions:
1) First we define a function $f(m,l)$ that counts decreasing sequences of length $l$, with upper bound $m$. We see that $f(m,l) = C(m,l)$ since it corresponds to picking an $l$-element subset from a $m$-element set.
2) We see, that there are $l+1$ ways to form tuples $(a,b)$ s.t. $a+b = l$. These are $(0,l), (1,l-1), \ldots, (l-1,1), (l,0)$.
Using this, we count
$g(l) = \sum_{j=0}^l \sum_{i=1}^n f(i-1, j) \cdot f(i-1,l-j)$
to get the number of sequences of total length $l+1$ (the maximum value, $i$, is spliced into the middle the sequences).
3) Now sequences should be "padded" with repetitive elements. This is essentially the problem of $k$ indistinguishable balls in $l$ distinguishable boxes.
This is summed for all $l$.
$\sum_{l=1}^k C(l + k - 1, k) \cdot g(l-1)$
gives the number of unimodular sequences.
Closed formula might exist, and Maple/Mathematica might even serve it directly to you :-).