Django InlineModelAdmin: Show partially an inline model and link to the complete model

UPDATE:
Since Django 1.8, this is built in.

See this answer and the official documentation.

OLD ANSWER:

At the end I found a simple solution.

I create a new template called linked.html that is a copy of tabular.html and I added this code to create the link.

{% if inline_admin_form.original.pk %}
          <td class="{{ field.field.name }}">
              <a href="https://stackoverflow.com/admin/{{ app_label }}/{{ inline_admin_formset.opts.admin_model_path }}/{{ inline_admin_form.original.pk }}/">Full record</a>
          </td>
{% endif %}

then I created a new model LinkedInline inheriting InlineModelAdmin

#override of the InlineModelAdmin to support the link in the tabular inline
class LinkedInline(admin.options.InlineModelAdmin):
    template = "admin/linked.html"
    admin_model_path = None

    def __init__(self, *args):
        super(LinkedInline, self).__init__(*args)
        if self.admin_model_path is None:
            self.admin_model_path = self.model.__name__.lower()

Then when I define a new inline, I have only to use my LinkedInline instead of the normal InlineModelAdmin.

I hope it can be useful for other people.

Giovanni