Your results do not appear correct. This is easy to see, without any calculation, because in your table, your $E[X_{(1)}]$ increases with sample size $n$; plainly, the expected value of the sample minimum must get smaller (i.e. become more negative) as the sample size $n$ gets larger.
The problem is conceptually quite easy.
In brief: if $X$ ~ $N(0,1)$ with pdf $f(x)$:
... then the pdf of the 1st order statistic (in a sample of size $n$) is:
... obtained here using the OrderStat function in mathStatica, with domain of support:
Then, $E[X_{(1)}]$, for $n = 1,2,3$ can be easily obtained exactly as:
The exact $n = 3$ case is approximately $-0.846284$, which is obviously different to your workings of -1.06 (line 1 of your Table), so it seems clear something is wrong with your workings (or perhaps my understanding of what you are seeking).
For $n \ge 4$, obtaining closed-form solutions is more tricky, but even if symbolic integration proves difficult, we can always use numerical integration (to arbitrary precision if desired). This is really very easy ... here, for instance, is $E[X_{(1)}]$, for sample size $n = 1$ to 14, using Mathematica:
sol = Table[NIntegrate[x g, {x, -Infinity, Infinity}], {n, 1, 14}]
{0., -0.56419, -0.846284, -1.02938, -1.16296, -1.26721, -1.35218, -1.4236, -1.48501,
-1.53875, -1.58644, -1.62923, -1.66799, -1.70338}
All done. These values are obviously very different to those in your table (right hand column).
To consider the more general case of a $N(\mu, \sigma^2)$ parent, proceed exactly as above, starting with the general Normal pdf.
What is the intended use of the result? That bears on what form of answer is needed, to include whether a stochastic (Monte Carlo) simulation approach might be adequate, And even the bigger picture matter of is this problem necessary to solve, and did someone come up with this problem as a way of solving a higher level problem, and there might be a better approach to the higher level problem which doesn't require this.
Here is a stochastic (Monte Carlo) simulation solution in MATLAB.
a = 1; b = 2; c = 3; d = 4; k = -1; % Made up values for illustrative purpose
n = 1e8; % Number of replications
mux = 10; sigmax = 4; sigmay = 7; % Made up values for illustrative purposes
X = mux + sigmax * randn(n,1); Y = sigmay * randn(n,1); Y1 = a + b + c + d * Y;
success_index = exp(X).*Y1 > 0; % replications in which condition is true
num_success = sum(success_index);
Cond_Sample = exp(X(success_index)) .* Y1(success_index) + k;
disp([num_success mean(Cond_Sample) std(Cond_Sample)/sqrt(num_success)])
1.0e+09 *
0.058475265000000 1.502775087443930 0.057342191058931
Best Answer
So that this does not remain an unanswered question, your calculations are correct, and there is no easier way of doing the calculations.