How can we depict a turned shape simply and nicely using ascii characters in 3D space?

I suggest this method :

Instead of a circle we use a square rotated by 45° for the cutting section so that we only need '/' and '' characters to draw it.

 /
/ 
 /
 /

And we use '_' character for the profiles: upper, lower and median.

 _________ 
 / 
/ ________ 
 / / 
 /________/ 

Isn't it Turning complete? Well, if you agree, write a full program or a function taking an unsigned integer value N , representing a number of steps , producing a 3d shape as described below.

The profile of this turned shape has the form of a stair step curve raising from 0 to N steps and lowering back to 0 where each step is 2 char('/') high and 5 char long ('_').

Maybe some examples describes it more clearly.

For N = 0 you may output nothing but it's not mandatory to handle it.

 N = 1

 _____ 
 / 
 / ____ 
 / / 
 /____/ 
.

 N = 2

 _____
 / 
 ___/_ __ 
 / 
 / ____ _____ 
 / / / / / 
 /____/ / /_/ 
 / / 
 /____/ 
.

 N = 3
 _____ 
 / 
 ___/_ __ 
 / 
 ___/_ __ 
 / 
 / ____ ____ ______ 
 / / / / / / / / 
 /____/ / / / / /_/ 
 / / / / / 
 /____/ / /_/ 
 / / 
 /____/ 

Rules :

- Margins are not specified.

- Standard loopholes are forbidden.

- Standard input/output methods.

- Shortest answer in bytes wins.

Charcoal, 70 69 bytes

ＮθＦθ«Ｊ⊕×⁷⊕ι⁰≔⊗⊕ιι↙ι↑←×⁵_Ｐ↖²→↗ι×⁴_↖ι←×⁵_↓Ｐ↙²→↘ιＪ⁻×⁹θ⊗ι⊖ι__↗ι←Ｐ←__↖ι←__

Try it online! Link is to verbose version of code. Explanation:

ＮθＦθ«

Input N and loop that many times.

Ｊ⊕×⁷⊕ι⁰

Jump to the (middle) right corner of the front slice.

≔⊗⊕ιι

Get the size of the slice.

↙ι↑←×⁵_Ｐ↖²→↗ι×⁴_↖ι←×⁵_↓Ｐ↙²→↘ι

Draw the front slice.

Ｊ⁻×⁹θ⊗ι⊖ι

Jump to the bottom (that we can see) of the back slice.

__↗ι←Ｐ←__↖ι←__

Draw the back slice.

JavaScript (ES8),  322 282 278  272 bytes

Builds the output line by line.

f=(n,y=0,Y=(k=y>2*n)?4*n-y:y,S='___/_25__,/\ \25,\/____/14_/,\ 14,\/____/,_____, \____\, / /, \ \,_\, /, \'.split`,`)=>~Y?(y-2*n?''.padEnd(n*5-(Y>>1)*5-3-y%2-k)+S[y?Y?k*2+y%2:4:5]:'/ 03').replace(/d/g,n=>S[+n+6].repeat(Y-(n>2||-k)>>1))+`
`+f(n,y+1):''

Try it online!

How?

For each row $0 le y le 4n$, we define:

$$k=cases
0,&textif $y le 2n$\
1,&textif $y > 2n$
$$
$$Y=cases
y,&textif $k=0$\
4n-y,&textif $k=1$
$$

Each row is first converted into a main pattern. A main pattern may contain digits: they are placeholders for repeated sub-patterns that are expended afterwards.

The middle row (when $y=2n$) is a special case which is processed separately. For all other rows, we compute the main pattern ID $p$ with:

$$p=cases
5,&textif $y=0$\
4,&textif $y neq 0, Y=0$\
2k+(y bmod 2),&textotherwise\
$$

There are no leading spaces for the middle row. For all other rows, the number $s$ of leading spaces is given by:

$$s=5n-5leftlfloorfracY2rightrfloor-3-(ybmod 2)-k$$

There are inner sub-patterns (marked with a digit $le2$) and outer sub-patterns (marked with a digit $>2$), which are repeated $n_1$ and $n_2$ times respectively:

$$n_1=leftlfloorfracY+k2rightrfloor\
n_2=leftlfloorfracY-12rightrfloor$$

The above formulae apply to the middle row as well, but are irrelevant for the first and last rows, which do not have sub-patterns.

Below is what we get for $n=3$:

 y Y k | p | s | before .replace() | n1 | n2 | after .replace()
--------+-----+-----+----------------------+----+----+-----------------------------
 0 0 0 | 5 | 12 | ............_____ | 0 | -1 | ............_____ 
 1 1 0 | 1 | 11 | .........../ 25 | 0 | 0 | .........../ 
 2 2 0 | 0 | 7 | .......___/_25__ | 1 | 0 | .......___/_ __ 
 3 3 0 | 1 | 6 | ....../ 25 | 1 | 1 | ....../ 
 4 4 0 | 0 | 2 | ..___/_25__ | 2 | 1 | ..___/_ __ 
 5 5 0 | 1 | 1 | ./ 25 | 2 | 2 | ./ 
 6 6 0 | n/a | n/a | / 03 | 3 | 2 | / ____ ____ ______
 7 5 1 | 3 | 0 | 14 | 3 | 2 | / / / / / / / /
 8 4 1 | 2 | 1 | ./____/14_/ | 2 | 1 | ./____/ / / / / /_/ 
 9 3 1 | 3 | 5 | ..... 14 | 2 | 1 | ..... / / / / / 
 10 2 1 | 2 | 6 | ....../____/14_/ | 1 | 0 | ....../____/ / /_/ 
 11 1 1 | 3 | 10 | .......... 14 | 1 | 0 | .......... / / 
 12 0 1 | 4 | 11 | .........../____/ | 0 | -1 | .........../____/