How to define new command for \{e_1,\cdots, e_n\}?

The following is an elementary implementation of what you're after:

\documentclass{article}

\newcommand{\set}[3]{%
  \{ #1#2{1},
  \ifnum\pdfstrcmp{#3}{3}=0
    #1#2 2
  \else
    \ldots
  \fi
  , #1#2{#3} \}
}

\begin{document}

$\set{a}{^}{3}$

$\set{e}{_}{n}$

\end{document}

We condition on what is presented as argument #3 using \pdfstrcmp. \pdfstrcmp{<strA>}{<strB>} does a string comparison between <strA> and <strB>, returning -1 | 0 | 1 if <strA> < | = | > <strB> (lexicographically).

The only difference between the two sets is the way the second term in the sequence is being set, so those elements are set without consideration of the condition. No consideration is added for n=1 or n=2.

---

A supplementary interface using a single argument:

\documentclass{article}

\newcommand{\set}[1]{\setaux#1\relax}
\def\setaux#1#2#3\relax{%
  \{ {#1}#2 1,
  \ifnum\pdfstrcmp{#3}{3}=0
    {#1}#2 2
  \else
    \ldots
  \fi
  , {#1}#2{#3} \}
}

\begin{document}

$\set{a^3}$

$\set{e_n}$

$\set{\alpha^{22}}$

\end{document}

With expl3:

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\set}{mmm}
 {
  \lbrace
  \str_case:nnF { #3 }
   {
    {1}{#1#2{1}}
    {2}{#1#2{1},#1#2{2}}
    {3}{#1#2{1},#1#2{2},#1#2{3}}
   }
   { #1#2{1},\dots,#1#2{#3} }
   \rbrace
 }
\ExplSyntaxOff

\begin{document}

$\set{a}{^}{3}$

$\set{a}{^}{n}$

$\set{e}{_}{2}$

$\set{e}{_}{k}$

\end{document}

Alternative syntax and implementation: the call is \set[<opt>]{<letter><^|_>} where <opt> is a number or a generic subscript. In case of a number, the sequence is spelt out fully. In the unusual case you want a number, but you prefer the dots, use [{{9}}] like in the last example. The mandatory argument has the form a^ or a_.

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\set}{O{n}m}
 {
  \lbrace
  \regex_match:nnTF { \A \d+ \Z } { #1 }
   {
    #2{1}
    \int_step_inline:nnnn { 2 } { 1 } { #1 }
     {
      ,#2{##1}
     }
   }
   {
    #2{1},\dots,#2{#1}
   }
  \rbrace
 }
\ExplSyntaxOff

\begin{document}

$\set[3]{a^}$ $\set{a^}$ $\set[k]{a^}$

$\set[2]{e_}$ $\set[4]{e_}$ $\set[{{9}}]{e_}$

\end{document}

This solution uses etoolbox and pgffor. Credits for @TH.'s answer on Testing for number.

This implementation also handles \mycmd{a}{^}{K}, where K is any positive integer.

\documentclass{article}

\usepackage{etoolbox}
\usepackage{pgffor}
\makeatletter
\newcommand\ifnumber[1]{%
    \begingroup
    \edef\temp{#1}%
    \expandafter\ifstrempty\expandafter{\temp}
    {\endgroup\@secondoftwo}
    {\expandafter\ifnumber@i\temp\@nnil}%
}
\def\ifnumber@i#1#2\@nnil{%
    \if-#1%
    \ifstrempty{#2}
    {\def\temp{X}}
    {\def\temp{#2}}%
    \else
    \def\temp{#1#2}%
    \fi
    \afterassignment\ifnumhelper
    \count@0\temp\relax\@nnil
    \endgroup
}

\def\numrelax{\relax}%
\def\ifnumhelper#1\@nnil{%
    \def\temp{#1}%
    \ifx\temp\numrelax
    \aftergroup\@firstoftwo
    \else
    \aftergroup\@secondoftwo
    \fi
}
\makeatother

\newcommand{\mycmd}[3]{%
        \ifnumber{#3}{%
                \ensuremath{%
                    \foreach \n in {1,...,#3}{#1#2{\n}\ifnum \n < #3 {,} \fi}
            }
        }{%
        \ensuremath{#1#2 1,\ldots,#1#2{#3}}%
    }%
}

\begin{document}

\mycmd{1}{^}{4}

\mycmd{a}{^}{3}

\mycmd{1}{^}{1}

\mycmd{a}{^}{n}

\mycmd{b}{_}{6}

\mycmd{3}{_}{n}

\end{document}