[Solved]: Is the universe problem for one-counter automata with restricted alphabet size undecidable?

It must be undecidable for an alphabet with two symbols. It is possible to code any alphabet into two letters, e.g., map 16 symbols to the length 4 binary strings $aaaa, aaab, dots, bbbb$. Then equality to $Sigma^*$ is equivalent to equality to all possible codes for strings. In the 16 letter example this means equality to all strings of a multiple of four letters. Clearly that is not universality. That is obtained by adding those binary strings that are not coding. That is a regular set and can be generated by a one counter automaton. The same explanation, with $LaTeX$ for those who appreciate it. Assume universality is undecidable for $Sigma$. Let $h: Sigma^* to {0,1}^*$ be an injective morphism. Now $L = Sigma^*$ iff $h(L) = h(Sigma^*)$. This in turn is equivalent to $h(L) cup R = {0,1}^*$ where $R$ is the (fixed) regular language ${0,1}^* – h(Sigma^*)$. Hence we cannot decide whether the binary one counter language $h(L) cup R$ is universal. Note that language is one counter as the family is closed under morphisms and union (with regular languages). As you state “I think that” I can also confirm the question is decidable for a one letter alphabet. It is decidable for push-down automata (hence context-free languages) as one letter CFL are (effectively) equivalent to regular languages.