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):'' 

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How?

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

$$k=\cases{ 0,&\text{if $y \le 2n$}\\ 1,&\text{if $y > 2n$} }$$ $$Y=\cases{ y,&\text{if $k=0$}\\ 4n-y,&\text{if $k=1$} }$$

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:

$$p=\cases{ 5,&\text{if $y=0$}\\ 4,&\text{if $y \neq 0, Y=0$}\\ 2k+(y \bmod 2),&\text{otherwise}\\ }$$

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

$$s=5n-5\left\lfloor\frac{Y}{2}\right\rfloor-3-(y\bmod 2)-k$$

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:

$$n_1=\left\lfloor\frac{Y+k}{2}\right\rfloor\\ n_2=\left\lfloor\frac{Y-1}{2}\right\rfloor$$

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 | ...........\/____/ 

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.